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Calculus

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Question 41Question →

Given that y=3x−1(x+2)2y = \frac{3x - 1}{(x + 2)^2}, what is the value of dydx\frac{dy}{dx} at x=1x = 1?

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Answer: 527\frac{5}{27}

Answer

The value of dydx\frac{dy}{dx} at x=1x = 1 is 527\frac{5}{27}.
Applying the quotient rule dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} with u=3x−1u = 3x - 1 and v=(x+2)2v = (x + 2)^2 yields dydx=8−3x(x+2)3\frac{dy}{dx} = \frac{8 - 3x}{(x + 2)^3}. Substituting x=1x = 1 results in 8−3(1)(1+2)3=527\frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{27}.

Step-by-Step Solution

1
Identify the numerator u(x)u(x) and denominator v(x)v(x) for the quotient rule.
Let u=3x−1u = 3x - 1 and v=(x+2)2v = (x + 2)^2.
The given function y=uvy = \frac{u}{v} requires the quotient rule dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
2
Differentiate u(x)u(x) and v(x)v(x) with respect to xx.
u′=3u' = 3 and, using the chain rule, v′=2(x+2)(1)=2(x+2)v' = 2(x + 2)(1) = 2(x + 2).
The derivative of the inner term (x+2)(x + 2) is 11, giving v′=2(x+2)v' = 2(x + 2).
3
Substitute u,v,u′,v′u, v, u', v' into the quotient rule formula and simplify.
dydx=3(x+2)2−(3x−1)⋅2(x+2)(x+2)4=(x+2)[3(x+2)−2(3x−1)](x+2)4=3x+6−6x+2(x+2)3=8−3x(x+2)3\frac{dy}{dx} = \frac{3(x + 2)^2 - (3x - 1) \cdot 2(x + 2)}{(x + 2)^4} = \frac{(x + 2)[3(x + 2) - 2(3x - 1)]}{(x + 2)^4} = \frac{3x + 6 - 6x + 2}{(x + 2)^3} = \frac{8 - 3x}{(x + 2)^3}.
Factoring out (x+2)(x + 2) simplifies the algebraic expression.
4
Evaluate the derivative at x=1x = 1.
dydx∣x=1=8−3(1)(1+2)3=533=527\frac{dy}{dx}\Big|_{x=1} = \frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{3^3} = \frac{5}{27}.
Substituting x=1x = 1 gives the final numerical derivative value.

Key Concept

Quotient and Chain Rules of Differentiation
Question 42Question →

Given the function y=2x−3(x2+1)2y = \frac{2x - 3}{(x^2 + 1)^2}, find the numerical value of dydx\frac{dy}{dx} at x=1x = 1.

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Answer: 1

Answer

The numerical value of dydx\frac{dy}{dx} at x=1x = 1 is 1.
Applying the quotient rule dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} alongside the chain rule for the denominator yields dydx=2(x2+1)2−(2x−3)⋅4x(x2+1)(x2+1)4\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}. Substituting x=1x = 1 evaluates to 2(4)−(−1)(8)16=1616=1\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1.

Step-by-Step Solution

1
Identify the numerator and denominator functions
Let u(x)=2x−3u(x) = 2x - 3 and v(x)=(x2+1)2v(x) = (x^2 + 1)^2.
The given function is structured as a quotient y=uvy = \frac{u}{v}, requiring the quotient rule.
2
Find the derivatives u′(x)u'(x) and v′(x)v'(x)
u′(x)=2u'(x) = 2 and v′(x)=2(x2+1)⋅2x=4x(x2+1)v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1).
Differentiating u(x)u(x) follows standard polynomial rules; v(x)v(x) requires the chain rule.
3
Evaluate u(1),u′(1),v(1),u(1), u'(1), v(1), and v′(1)v'(1) at x=1x = 1
u(1)=−1u(1) = -1, u′(1)=2u'(1) = 2, v(1)=4v(1) = 4, and v′(1)=8v'(1) = 8.
Evaluating components before substitution simplifies the arithmetic.
4
Apply the quotient rule formula dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} at x=1x = 1
\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.
Substitute the calculated component values into the quotient rule formula.

Key Concept

Quotient Rule and Chain Rule of Differentiation
Estimated Time:1m 30s
Question 43Question →

Given the function y=x2+2x(3x−1)2y = \frac{x^2 + 2x}{(3x - 1)^2}, what is the value of dydx\frac{dy}{dx} at x=1x = 1?

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Answer: −54-\frac{5}{4}

Answer

−54-\frac{5}{4}
Applying the quotient rule dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2} with u=x2+2xu = x^2 + 2x (u′=2x+2u' = 2x + 2) and v=(3x−1)2v = (3x - 1)^2 (v′=6(3x−1)v' = 6(3x - 1)) yields dydx=16−3616=−54\frac{dy}{dx} = \frac{16 - 36}{16} = -\frac{5}{4} at x=1x = 1.

Step-by-Step Solution

1
Identify the numerator and denominator functions for the quotient rule y=uvy = \frac{u}{v}.
u=x2+2xu = x^2 + 2x and v=(3x−1)2v = (3x - 1)^2.
The function is structured as a quotient of two algebraic expressions.
2
Differentiate uu and vv with respect to xx.
dudx=2x+2\frac{du}{dx} = 2x + 2 and dvdx=2(3x−1)⋅3=6(3x−1)\frac{dv}{dx} = 2(3x - 1) \cdot 3 = 6(3x - 1).
Use the power rule for uu and the chain rule for vv.
3
Apply the quotient rule formula dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.
dydx=(3x−1)2(2x+2)−(x2+2x)⋅6(3x−1)(3x−1)4\frac{dy}{dx} = \frac{(3x - 1)^2 (2x + 2) - (x^2 + 2x) \cdot 6(3x - 1)}{(3x - 1)^4}.
The quotient rule formula combines the expressions and their derivatives.
4
Substitute x=1x = 1 into the derivative expression and simplify.
dydx=(2)2(4)−(3)⋅6(2)(2)4=16−3616=−2016=−54\frac{dy}{dx} = \frac{(2)^2 (4) - (3) \cdot 6(2)}{(2)^4} = \frac{16 - 36}{16} = -\frac{20}{16} = -\frac{5}{4}.
Evaluating at x=1x = 1 yields the numerical derivative value.

Key Concept

Quotient Rule and Chain Rule of Differentiation
Question 44Question →

What is the indefinite integral ∫(3x2−5sin⁡(5x)+2) dx\int (3x^2 - 5\sin(5x) + 2) \, dx?

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Answer: x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C

Answer

x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C
Integrating term-by-term yields ∫3x2dx=x3\int 3x^2 dx = x^3, ∫−5sin⁡(5x)dx=cos⁡(5x)\int -5\sin(5x) dx = \cos(5x), and ∫2dx=2x\int 2 dx = 2x. Adding the arbitrary constant CC produces x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C.

Step-by-Step Solution

1
Integrate the polynomial term 3x23x^2 using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}
∫3x2dx=3⋅x33=x3\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3
Applying the power rule for integration.
2
Integrate the trigonometric term −5sin⁡(5x)-5\sin(5x) using ∫sin⁡(kx)dx=−1kcos⁡(kx)\int \sin(kx) dx = -\frac{1}{k}\cos(kx)
∫−5sin⁡(5x)dx=−5(−15cos⁡(5x))=cos⁡(5x)\int -5\sin(5x) dx = -5 \left(-\frac{1}{5}\cos(5x)\right) = \cos(5x)
Integration of the sine function reverses differentiation with a positive sign change for negative sine.
3
Integrate the constant term 22 and append the constant of integration CC
∫2dx=2x\int 2 dx = 2x, giving total antiderivative x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C
Indefinite integrals always require an arbitrary constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 45Question →

What is the x-coordinate of the maximum stationary point of the curve y=sin⁡x+cos⁡xy = \sin x + \cos x in the interval 0≤x≤π0 \le x \le \pi?

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Answer: π4\frac{\pi}{4}

Answer

The x-coordinate of the maximum stationary point is π4\frac{\pi}{4}.
To find stationary points, we differentiate y=sin⁡x+cos⁡xy = \sin x + \cos x to get dydx=cos⁡x−sin⁡x\frac{dy}{dx} = \cos x - \sin x. Setting dydx=0\frac{dy}{dx} = 0 gives sin⁡x=cos⁡x\sin x = \cos x, or tan⁡x=1\tan x = 1. Within 0≤x≤π0 \le x \le \pi, the solution is x=π4x = \frac{\pi}{4}. Evaluating the second derivative d2ydx2=−sin⁡x−cos⁡x\frac{d^2y}{dx^2} = -\sin x - \cos x at x=π4x = \frac{\pi}{4} yields −2<0-\sqrt{2} < 0, confirming a local maximum.

Step-by-Step Solution

1
Differentiate the function y=sin⁡x+cos⁡xy = \sin x + \cos x with respect to xx.
dydx=cos⁡x−sin⁡x\frac{dy}{dx} = \cos x - \sin x
Stationary points occur where the first derivative dydx=0\frac{dy}{dx} = 0.
2
Set the first derivative to zero and solve for xx in the interval 0≤x≤π0 \le x \le \pi.
\cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}
Dividing both sides by cos⁡x\cos x gives tan⁡x=1\tan x = 1, which has the solution x=π4x = \frac{\pi}{4} in the given domain.
3
Verify the nature of the stationary point using the second derivative test.
\frac{d^2y}{dx^2} = -\sin x - \cos x \implies \left.\frac{d^2y}{dx^2}\right|_{x=\frac{\pi}{4}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0
A negative second derivative confirms that x=π4x = \frac{\pi}{4} is a local maximum point.

Key Concept

Finding stationary points and determining their nature using derivatives of trigonometric functions.
Question 46Question →

Find the positive constant pp such that the value of the definite integral ∫0p(3x2+2x) dx\int_{0}^{p} (3x^2 + 2x) \, dx is equal to 1212.

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Answer: 2

Answer

The positive constant pp is equal to 22.
Evaluating the definite integral ∫0p(3x2+2x) dx\int_{0}^{p} (3x^2 + 2x) \, dx gives p3+p2p^3 + p^2. Setting this expression equal to 1212 forms the cubic equation p3+p2−12=0p^3 + p^2 - 12 = 0. Solving for the positive real solution yields p=2p = 2, since 23+22=8+4=122^3 + 2^2 = 8 + 4 = 12.

Step-by-Step Solution

1
Find the antiderivative of the integrand
\int (3x^2 + 2x) \, dx = x^3 + x^2 + C
Applying the power rule for integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} term by term gives 3x33+2x22=x3+x2\frac{3x^3}{3} + \frac{2x^2}{2} = x^3 + x^2.
2
Evaluate the antiderivative at the limits 00 and pp
\left[ x^3 + x^2 \right]_0^p = (p^3 + p^2) - (0^3 + 0^2) = p^3 + p^2
Using the Fundamental Theorem of Calculus: substitute the upper limit pp and subtract the value at the lower limit 00.
3
Set up the polynomial equation using the given integral value
p^3 + p^2 = 12 \implies p^3 + p^2 - 12 = 0
The problem states that the definite integral equals 12.
4
Solve for the positive real value of pp
p = 2
Testing p=2p=2: 23+22=8+4=122^3 + 2^2 = 8 + 4 = 12. Factoring gives (p−2)(p2+3p+6)=0(p-2)(p^2 + 3p + 6) = 0, where the quadratic factor has a negative discriminant 32−4(1)(6)=−15<03^2 - 4(1)(6) = -15 < 0, making p=2p=2 the only real solution.

Key Concept

Definite integration with an unknown limit of integration
Question 47Question →

What is the value of the definite integral ∫0π3cos⁡(2x) dx\int_{0}^{\frac{\pi}{3}} \cos(2x) \, dx?

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Answer: 34\frac{\sqrt{3}}{4}

Answer

34\frac{\sqrt{3}}{4}
The antiderivative of cos⁡(2x)\cos(2x) is 12sin⁡(2x)\frac{1}{2}\sin(2x). Substituting the limits gives 12sin⁡(2π3)−12sin⁡(0)=12(32)−0=34\frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}, which is correct.

Step-by-Step Solution

1
Find the antiderivative of cos⁡(2x)\cos(2x)
∫cos⁡(2x) dx=12sin⁡(2x)+C\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C
Using standard trigonometric integration rules, ∫cos⁡(kx) dx=1ksin⁡(kx)+C\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C.
2
Substitute the upper limit x=π3x = \frac{\pi}{3} and lower limit x=0x = 0
[12sin⁡(2x)]0π3=12sin⁡(2π3)−12sin⁡(0)\left[\frac{1}{2}\sin(2x)\right]_{0}^{\frac{\pi}{3}} = \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0)
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x)dx = F(b) - F(a).
3
Evaluate the trigonometric values and simplify
12(32)−0=34\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}
sin⁡(2π3)=sin⁡(120∘)=32\sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2} and sin⁡(0)=0\sin(0) = 0.

Key Concept

Definite Integration of Trigonometric Functions
Question 48Question →

What is the value of the definite integral ∫12(4x3−3x2+2) dx\int_{1}^{2} (4x^3 - 3x^2 + 2) \, dx?

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Answer: 10

Answer

10
Integrating each term yields F(x)=x4−x3+2xF(x) = x^4 - x^3 + 2x. Substituting the upper boundary gives F(2)=12F(2) = 12 and the lower boundary gives F(1)=2F(1) = 2. Calculating F(2)−F(1)F(2) - F(1) gives 12−2=1012 - 2 = 10, which is the exact value of the definite integral.

Step-by-Step Solution

1
Find the indefinite antiderivative of the integrand 4x3−3x2+24x^3 - 3x^2 + 2.
F(x)=∫(4x3−3x2+2) dx=x4−x3+2xF(x) = \int (4x^3 - 3x^2 + 2) \, dx = x^4 - x^3 + 2x
Apply the power rule of integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each term.
2
Evaluate the antiderivative at the upper limit x=2x = 2.
F(2)=(2)4−(2)3+2(2)=16−8+4=12F(2) = (2)^4 - (2)^3 + 2(2) = 16 - 8 + 4 = 12
Substitute x=2x = 2 into F(x)F(x).
3
Evaluate the antiderivative at the lower limit x=1x = 1.
F(1)=(1)4−(1)3+2(1)=1−1+2=2F(1) = (1)^4 - (1)^3 + 2(1) = 1 - 1 + 2 = 2
Substitute x=1x = 1 into F(x)F(x).
4
Subtract the lower limit value from the upper limit value.
F(2)−F(1)=12−2=10F(2) - F(1) = 12 - 2 = 10
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a).

Key Concept

Definite Integration of Polynomial Functions
Estimated Time:1m 30s
Question 49Question →

Find the value of the definite integral ∫12(x+1x2)dx\int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx.

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Answer: 2

Answer

The value of the definite integral is 2.
Integrating x+x−2x + x^{-2} gives x22−1x\frac{x^2}{2} - \frac{1}{x}. Evaluating from x=1x=1 to x=2x=2 yields (2−12)−(12−1)=32−(−12)=2\left(2 - \frac{1}{2}\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2.

Step-by-Step Solution

1
Find the antiderivative of f(x)=x+x−2f(x) = x + x^{-2}
F(x)=x22−1x+CF(x) = \frac{x^2}{2} - \frac{1}{x} + C
Apply the power rule of integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} for n≠−1n \neq -1.
2
Evaluate F(x)F(x) at the upper boundary x=2x = 2
F(2)=222−12=2−12=32F(2) = \frac{2^2}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}
Substitute x=2x = 2 into the antiderivative.
3
Evaluate F(x)F(x) at the lower boundary x=1x = 1
F(1)=122−11=12−1=−12F(1) = \frac{1^2}{2} - \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}
Substitute x=1x = 1 into the antiderivative.
4
Calculate F(2)−F(1)F(2) - F(1)
32−(−12)=32+12=2\frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a).

Key Concept

Definite Integration using the Power Rule
Question 50Question →

What is the yy-intercept of the normal line to the curve y=2x−1x+1y = \frac{2x - 1}{x + 1} at the point where the curve crosses the xx-axis?

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Answer: 38\frac{3}{8}

Answer

The yy-intercept of the normal line is 38\frac{3}{8}.
To find the yy-intercept of the normal line, set y=0y = 0 to find the point of contact on the xx-axis, which gives (12,0)\left(\frac{1}{2}, 0\right). Differentiating y=2x−1x+1y = \frac{2x - 1}{x + 1} gives dydx=3(x+1)2\frac{dy}{dx} = \frac{3}{(x + 1)^2}. At x=12x = \frac{1}{2}, the tangent slope is 43\frac{4}{3}, making the normal slope −34-\frac{3}{4}. The equation of the normal line is y=−34x+38y = -\frac{3}{4}x + \frac{3}{8}, so its yy-intercept is 38\frac{3}{8}.

Step-by-Step Solution

1
Find the point of intersection of the curve with the xx-axis
Set y=0  ⟹  2x−1x+1=0  ⟹  2x−1=0  ⟹  x=12y = 0 \implies \frac{2x - 1}{x + 1} = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}. The point is (12,0)\left(\frac{1}{2}, 0\right).
The normal line is drawn at the point where the curve crosses the xx-axis.
2
Differentiate y=2x−1x+1y = \frac{2x - 1}{x + 1} using the quotient rule
\(\frac{dy}{dx} = \frac{2(x + 1) - (2x - 1)(1)}{(x + 1)^2} = \frac{2x + 2 - 2x + 1}{(x + 1)^2} = \frac{3}{(x + 1)^2}\)
The derivative provides the gradient function of the tangent line to the curve.
3
Evaluate the gradient of the tangent and normal lines at x=12x = \frac{1}{2}
Tangent gradient mt=3(12+1)2=394=43m_t = \frac{3}{\left(\frac{1}{2} + 1\right)^2} = \frac{3}{\frac{9}{4}} = \frac{4}{3}. Normal gradient mn=−1mt=−34m_n = -\frac{1}{m_t} = -\frac{3}{4}.
The normal line is perpendicular to the tangent line at the point of contact.
4
Determine the equation of the normal line and evaluate its yy-intercept
Using y−0=−34(x−12)  ⟹  y=−34x+38y - 0 = -\frac{3}{4}\left(x - \frac{1}{2}\right) \implies y = -\frac{3}{4}x + \frac{3}{8}. Setting x=0x = 0 gives y=38y = \frac{3}{8}.
The yy-intercept is the value of yy when x=0x = 0 on the line.

Key Concept

Tangents and Normals to Curves
Question 51Question →

Using differentiation from first principles, evaluate the value of the derivative dydx=lim⁡h→0f(x+h)−f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} for the cubic function f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 at the point x=3x = 3.

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Answer: 12

Answer

The derivative evaluated at x=3x = 3 is equal to 12.
Evaluating the definition of the derivative from first principles for f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 yields lim⁡h→0f(x+h)−f(x)h=6x2−18x+12\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 6x^2 - 18x + 12. Substituting x=3x = 3 gives 6(3)2−18(3)+12=54−54+12=126(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12.

Step-by-Step Solution

1
Set up the difference quotient definition from first principles
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Differentiation from first principles requires finding the limit of the average rate of change as the increment hh approaches zero.
2
Substitute (x+h)(x+h) into f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 and expand
f(x+h)=2x3+6x2h+6xh2+2h3−9x2−18xh−9h2+12x+12h−5f(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 9x^2 - 18xh - 9h^2 + 12x + 12h - 5
Expanding binomial terms (x+h)3(x+h)^3 and (x+h)2(x+h)^2 reveals all components involving hh.
3
Calculate f(x+h)−f(x)f(x+h) - f(x) and factor out hh
f(x+h)−f(x)=h(6x2+6xh+2h2−18x−9h+12)f(x+h) - f(x) = h(6x^2 + 6xh + 2h^2 - 18x - 9h + 12)
Terms independent of hh cancel out completely, isolating hh as a common factor.
4
Divide by hh and evaluate the limit as h→0h \to 0
f′(x)=6x2−18x+12f'(x) = 6x^2 - 18x + 12
Canceling hh resolves the 00\frac{0}{0} indeterminate form, allowing direct substitution of h=0h=0.
5
Substitute x=3x = 3 into f′(x)f'(x)
f′(3)=6(3)2−18(3)+12=12f'(3) = 6(3)^2 - 18(3) + 12 = 12
Evaluating at x=3x = 3 gives the numerical value of the instantaneous rate of change at that specific point.

Key Concept

Differentiation from First Principles
Question 52Question →

Given the function y=(x2+1)32x−3y = \frac{(x^2 + 1)^3}{2x - 3}, what is the numerical value of dydx\frac{dy}{dx} evaluated at x=2x = 2?

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Answer: 50

Answer

The numerical value of dydx\frac{dy}{dx} at x=2x = 2 is 50.
Applying the Quotient Rule dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} along with the Chain Rule to find u′(x)=6x(x2+1)2u'(x) = 6x(x^2 + 1)^2, and evaluating all components at x=2x = 2 gives u(2)=125u(2) = 125, u′(2)=300u'(2) = 300, v(2)=1v(2) = 1, and v′(2)=2v'(2) = 2. Substituting these values yields (300)(1)−(125)(2)12=50\frac{(300)(1) - (125)(2)}{1^2} = 50.

Step-by-Step Solution

1
Set up the Quotient Rule components
u(x)=(x2+1)3u(x) = (x^2 + 1)^3 and v(x)=2x−3v(x) = 2x - 3
The given expression is a quotient of two functions requiring dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
2
Differentiate u(x)u(x) using the Chain Rule and v(x)v(x) using basic power rules
u′(x)=6x(x2+1)2u'(x) = 6x(x^2 + 1)^2 and v′(x)=2v'(x) = 2
Differentiating the outer power 3 gives 3(x2+1)23(x^2 + 1)^2, and multiplying by the derivative of the inner function (2x)(2x) gives 6x(x2+1)26x(x^2 + 1)^2.
3
Evaluate u(2)u(2), u′(2)u'(2), v(2)v(2), and v′(2)v'(2)
u(2)=125u(2) = 125, u′(2)=300u'(2) = 300, v(2)=1v(2) = 1, and v′(2)=2v'(2) = 2
Substituting x=2x = 2 into each function and derivative simplifies calculation of the overall derivative.
4
Substitute values into the Quotient Rule formula
dydx∣x=2=(300)(1)−(125)(2)(1)2=50\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(2)}{(1)^2} = 50
Evaluating u′(2)v(2)−u(2)v′(2)[v(2)]2\frac{u'(2)v(2) - u(2)v'(2)}{[v(2)]^2} yields the exact numerical result.

Key Concept

Combined Application of Quotient Rule and Chain Rule
Question 53Question →

What is the indefinite integral ∫((3−2x)(4x+1)+8cos⁡(4x))dx\int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx?

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Answer: −83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Answer

−83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C
Expanding the product (3−2x)(4x+1)(3 - 2x)(4x + 1) gives −8x2+10x+3-8x^2 + 10x + 3. Integrating term-by-term yields ∫−8x2dx=−83x3\int -8x^2 dx = -\frac{8}{3}x^3, ∫10xdx=5x2\int 10x dx = 5x^2, ∫3dx=3x\int 3 dx = 3x, and ∫8cos⁡(4x)dx=84sin⁡(4x)=2sin⁡(4x)\int 8\cos(4x) dx = \frac{8}{4}\sin(4x) = 2\sin(4x). Adding the integration constant CC gives the complete result −83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C.

Step-by-Step Solution

1
Expand the polynomial product inside the integrand
(3−2x)(4x+1)=12x+3−8x2−2x=−8x2+10x+3(3 - 2x)(4x + 1) = 12x + 3 - 8x^2 - 2x = -8x^2 + 10x + 3
Expanding the expression allows term-by-term integration using standard rules.
2
Rewrite the full integrand
∫(−8x2+10x+3+8cos⁡(4x))dx\int \left( -8x^2 + 10x + 3 + 8\cos(4x) \right) dx
Substitute the expanded polynomial back into the integral expression.
3
Integrate each term individually
∫−8x2dx=−83x3\int -8x^2 dx = -\frac{8}{3}x^3, ∫10xdx=5x2\int 10x dx = 5x^2, ∫3dx=3x\int 3 dx = 3x, and ∫8cos⁡(4x)dx=8⋅sin⁡(4x)4=2sin⁡(4x)\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)
Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} and linear trigonometric rule ∫cos⁡(kx)dx=sin⁡(kx)k\int \cos(kx) dx = \frac{\sin(kx)}{k}.
4
Combine terms and append the constant of integration
−83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C
Indefinite integration requires adding an arbitrary constant CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Estimated Time:2m 0s
Question 54Question →

What is the equation of the normal line to the curve y=x3−3x+2y = x^3 - 3x + 2 at the point where x=2x = 2?

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Answer: x+9y−38=0x + 9y - 38 = 0

Answer

The equation of the normal line is x+9y−38=0x + 9y - 38 = 0.
Evaluating yy at x=2x = 2 gives the point (2,4)(2, 4). Differentiating y=x3−3x+2y = x^3 - 3x + 2 yields dydx=3x2−3\frac{dy}{dx} = 3x^2 - 3. At x=2x = 2, the tangent slope is mt=9m_t = 9, making the normal slope mn=−1/9m_n = -1/9. Using the point-slope formula y−4=−1/9(x−2)y - 4 = -1/9(x - 2) simplifies to x+9y−38=0x + 9y - 38 = 0.

Step-by-Step Solution

1
Find the yy-coordinate at x=2x = 2
y=(2)3−3(2)+2=8−6+2=4y = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4. The point on the curve is (2,4)(2, 4).
The point of tangency/normal line intersection must be determined on the curve.
2
Calculate the derivative dydx\frac{dy}{dx} to find the slope of the tangent line
dydx=3x2−3\frac{dy}{dx} = 3x^2 - 3. At x=2x = 2, dydx=3(2)2−3=12−3=9\frac{dy}{dx} = 3(2)^2 - 3 = 12 - 3 = 9.
The derivative evaluated at x=2x = 2 gives the gradient of the tangent line.
3
Find the slope of the normal line
mn=−1mt=−19m_n = -\frac{1}{m_t} = -\frac{1}{9}.
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.
4
Use the point-slope form to find the equation of the normal line
y−4=−19(x−2)  ⟹  9(y−4)=−(x−2)  ⟹  9y−36=−x+2  ⟹  x+9y−38=0y - 4 = -\frac{1}{9}(x - 2) \implies 9(y - 4) = -(x - 2) \implies 9y - 36 = -x + 2 \implies x + 9y - 38 = 0.
Substituting point (2,4)(2, 4) and slope −1/9-1/9 into y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) yields the standard line equation.

Key Concept

The slope of the normal line to a curve y=f(x)y = f(x) at (x1,y1)(x_1, y_1) is mn=−1f′(x1)m_n = -\frac{1}{f'(x_1)}.
Question 55Question →
What is the numerical value of the limit lim⁡x→4x−4x+5−3\lim_{x \to 4} \frac{x - 4}{\sqrt{x + 5} - 3}?
Show answer & explanation

Answer: 66

Answer

The value of the limit is 66.
Multiplying both the numerator and the denominator by the conjugate of the denominator (x+5+3)(\sqrt{x + 5} + 3) allows the factor (x−4)(x - 4) to cancel out, leaving x+5+3\sqrt{x + 5} + 3. Evaluating this expression as x→4x \to 4 yields 3+3=63 + 3 = 6.

Step-by-Step Solution

1
Check for direct substitution.
Substituting x=4x = 4 yields 4−44+5−3=03−3=00\frac{4 - 4}{\sqrt{4 + 5} - 3} = \frac{0}{3 - 3} = \frac{0}{0}, which is an indeterminate form.
Direct substitution gives 00\frac{0}{0}, requiring algebraic simplification such as rationalization.
2
Rationalize the denominator by multiplying the numerator and denominator by the conjugate (x+5+3)(\sqrt{x + 5} + 3).
lim⁡x→4(x−4)(x+5+3)(x+5−3)(x+5+3)=lim⁡x→4(x−4)(x+5+3)(x+5)−9\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(\sqrt{x + 5} - 3)(\sqrt{x + 5} + 3)} = \lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(x + 5) - 9}
Using the difference of squares formula (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2 eliminates the square root in the denominator.
3
Simplify the denominator and cancel out the common factor (x−4)(x - 4).
lim⁡x→4(x−4)(x+5+3)x−4=lim⁡x→4(x+5+3)\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{x - 4} = \lim_{x \to 4} (\sqrt{x + 5} + 3)
Since x≠4x \neq 4 when evaluating the limit, the indeterminate factor (x−4)(x - 4) cancels out.
4
Substitute x=4x = 4 into the simplified expression.
4+5+3=9+3=3+3=6\sqrt{4 + 5} + 3 = \sqrt{9} + 3 = 3 + 3 = 6
Evaluates the limit after removing the zero-denominator condition.

Key Concept

Limits of indeterminate forms 00\frac{0}{0} involving radicals (Rationalization Technique)
Estimated Time:1m 30s
Question 56Question →

If y=ln⁡(e3xcos⁡x)y = \ln(e^{3x} \cos x), what is dydx\frac{dy}{dx}?

Show answer & explanation

Answer: 3−tan⁡x3 - \tan x

Answer

3−tan⁡x3 - \tan x
Applying the product rule for logarithms converts y=ln⁡(e3xcos⁡x)y = \ln(e^{3x} \cos x) into y=3x+ln⁡(cos⁡x)y = 3x + \ln(\cos x). Differentiating 3x3x yields 33, and differentiating ln⁡(cos⁡x)\ln(\cos x) by chain rule yields −sin⁡xcos⁡x=−tan⁡x\frac{-\sin x}{\cos x} = -\tan x. Combining these gives 3−tan⁡x3 - \tan x.

Step-by-Step Solution

1
Simplify the logarithmic function using log laws
y=ln⁡(e3x)+ln⁡(cos⁡x)=3x+ln⁡(cos⁡x)y = \ln(e^{3x}) + \ln(\cos x) = 3x + \ln(\cos x)
Splitting the logarithm of a product simplifies the differentiation process.
2
Differentiate each term with respect to xx
\frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}(\ln(\cos x)) = 3 + \frac{1}{\cos x} \cdot (-\sin x)
The derivative of 3x3x is 33, and applying the chain rule to ln⁡(u)\ln(u) gives 1u⋅dudx\frac{1}{u} \cdot \frac{du}{dx} where u=cos⁡xu = \cos x.
3
Simplify the trigonometric ratio
\frac{dy}{dx} = 3 - \frac{\sin x}{\cos x} = 3 - \tan x
Replacing sin⁡xcos⁡x\frac{\sin x}{\cos x} with tan⁡x\tan x yields the simplified final derivative.

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Laws
Question 57Question →
Evaluate the algebraic limit:
lim⁡x→4x−2x2−16\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16}
What is the exact value of this limit?
Show answer & explanation

Answer: 132\frac{1}{32}

Answer

The exact value of the limit is 132\frac{1}{32}.
The limit presents an indeterminate form 00\frac{0}{0} at x=4x = 4. Factoring x2−16x^2 - 16 into (x−2)(x+2)(x+4)(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4) allows the factor (x−2)(\sqrt{x} - 2) to be cancelled from both the numerator and denominator. Substituting x=4x = 4 into the simplified expression 1(x+2)(x+4)\frac{1}{(\sqrt{x} + 2)(x + 4)} yields 1(2+2)(4+4)=132\frac{1}{(2+2)(4+4)} = \frac{1}{32}.

Step-by-Step Solution

1
Identify the form of the limit by direct substitution.
Substituting x=4x = 4 into 4−242−16\frac{\sqrt{4} - 2}{4^2 - 16} gives 00\frac{0}{0}, which is an indeterminate form.
Direct substitution yields 00\frac{0}{0}, requiring algebraic simplification.
2
Factor the denominator x2−16x^2 - 16.
x2−16=(x−4)(x+4)x^2 - 16 = (x - 4)(x + 4)
Use the difference of two squares identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
3
Factor (x−4)(x - 4) as a difference of squares involving square roots.
x−4=(x)2−22=(x−2)(x+2)x - 4 = (\sqrt{x})^2 - 2^2 = (\sqrt{x} - 2)(\sqrt{x} + 2)
This exposes the vanishing factor (x−2)(\sqrt{x} - 2) in the denominator.
4
Cancel the common factor (x−2)(\sqrt{x} - 2) and evaluate the limit.
lim⁡x→4x−2(x−2)(x+2)(x+4)=lim⁡x→41(x+2)(x+4)=1(4+2)(4+4)=14×8=132\lim_{x \to 4} \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)} = \lim_{x \to 4} \frac{1}{(\sqrt{x} + 2)(x + 4)} = \frac{1}{(\sqrt{4} + 2)(4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}
Cancelling the factor removing the 00\frac{0}{0} condition allows direct evaluation.

Key Concept

Resolution of indeterminate limits of the form 0/0 using algebraic factorization and conjugate radical identities.
Estimated Time:2m 0s
Question 58Question →

A curve has the equation y=13x3−2x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1. What is the positive xx-coordinate of the point on the curve where the tangent line is parallel to the line y=8x−5y = 8x - 5?

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Answer: 5

Answer

The positive xx-coordinate of the point where the tangent is parallel to the line is 55.
The gradient of the line y=8x−5y = 8x - 5 is 88. Differentiating y=13x3−2x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1 gives dydx=x2−4x+3\frac{dy}{dx} = x^2 - 4x + 3. Setting dydx=8\frac{dy}{dx} = 8 leads to x2−4x−5=0x^2 - 4x - 5 = 0, which factors as (x−5)(x+1)=0(x - 5)(x + 1) = 0. The solutions are x=5x = 5 and x=−1x = -1. Selecting the positive value gives x=5x = 5.

Step-by-Step Solution

1
Find the gradient (slope) of the given straight line.
The line equation is in slope-intercept form y=mx+cy = mx + c, where the slope m=8m = 8.
Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.
2
Differentiate the curve equation to find the gradient function dydx\frac{dy}{dx}.
dydx=ddx(13x3−2x2+3x+1)=x2−4x+3\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3.
The first derivative of a curve represents the gradient of the tangent at any point xx.
3
Equate the derivative to the slope of the line and solve for xx.
x2−4x+3=8  ⟹  x2−4x−5=0  ⟹  (x−5)(x+1)=0x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0. The roots are x=5x = 5 and x=−1x = -1.
Solving the quadratic equation gives all xx-values where the tangent line has a slope of 8.
4
Select the positive xx-coordinate as requested.
x=5x = 5.
The question specifically asks for the positive value among the solutions.

Key Concept

Finding points on a curve where the tangent is parallel to a given line
Question 59Question →

If y=(2x−1)3(x2+1)2y = (2x - 1)^3(x^2 + 1)^2, find the numerical value of dydx\frac{dy}{dx} at x=1x = 1.

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Answer: 32

Answer

The numerical value of dydx\frac{dy}{dx} evaluated at x=1x = 1 is 3232.
Using the product rule together with the chain rule for composite functions, we find u′(x)=6(2x−1)2u'(x) = 6(2x - 1)^2 and v′(x)=4x(x2+1)v'(x) = 4x(x^2 + 1). Evaluating at x=1x = 1 gives u(1)=1u(1)=1, u′(1)=6u'(1)=6, v(1)=4v(1)=4, and v′(1)=8v'(1)=8. Calculating dydx=u′(1)v(1)+u(1)v′(1)=6(4)+1(8)=32\frac{dy}{dx} = u'(1)v(1) + u(1)v'(1) = 6(4) + 1(8) = 32.

Step-by-Step Solution

1
Set up the product rule for y=u(x)v(x)y = u(x)v(x)
u(x)=(2x−1)3u(x) = (2x - 1)^3 and v(x)=(x2+1)2v(x) = (x^2 + 1)^2
The given function is a product of two composite expressions.
2
Differentiate u(x)u(x) using the chain rule
u′(x)=3(2x−1)2⋅2=6(2x−1)2u'(x) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2
Differentiating the outer power function and multiplying by the derivative of the inner function 2x−12x - 1.
3
Differentiate v(x)v(x) using the chain rule
v′(x)=2(x2+1)⋅2x=4x(x2+1)v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)
Differentiating the outer power function and multiplying by the derivative of the inner function x2+1x^2 + 1.
4
Apply the product rule formula dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) and evaluate at x=1x = 1
u(1)=1u(1) = 1, u′(1)=6u'(1) = 6, v(1)=4v(1) = 4, v′(1)=8v'(1) = 8, giving dydx∣x=1=(6)(4)+(1)(8)=32\frac{dy}{dx}\Big|_{x=1} = (6)(4) + (1)(8) = 32
Substituting x=1x = 1 into each individual term simplifies the arithmetic before combining.

Key Concept

Combined application of the Product Rule and Chain Rule
Question 60Question →

Evaluate the definite integral ∫02(3x2+4) dx\int_{0}^{2} (3x^2 + 4) \, dx.

Show answer & explanation

Answer: 16

Answer

The value of the definite integral is 1616.
Integrating 3x2+43x^2 + 4 with respect to xx gives the antiderivative F(x)=x3+4xF(x) = x^3 + 4x. Evaluating this antiderivative at the upper limit x=2x = 2 yields 23+4(2)=162^3 + 4(2) = 16, and at the lower limit x=0x = 0 yields 03+4(0)=00^3 + 4(0) = 0. Subtracting the lower boundary value from the upper boundary value gives 16−0=1616 - 0 = 16.

Step-by-Step Solution

1
Integrate the polynomial term by term
\int (3x^2 + 4) dx = x^3 + 4x
Apply the power rule of integration \int x^n dx = \frac{x^{n+1}}{n+1} and \int k dx = kx.
2
Apply the fundamental theorem of calculus with limits 0 and 2
[x^3 + 4x]_0^2 = (2^3 + 4(2)) - (0^3 + 4(0)) = 16 - 0 = 16
Evaluate F(b) - F(a) where F(x) is the antiderivative.

Key Concept

Definite Integral Evaluation using the Fundamental Theorem of Calculus
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