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Rules of Differentiation (Product, Quotient, and Chain Rules)

21 questions

All types21Multiple choice8Numeric13
Question 1Question →

Given the function y=(x2+1)33x−5y = \frac{(x^2 + 1)^3}{3x - 5}, calculate the value of dydx\frac{dy}{dx} at x=2x = 2.

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Answer: -75

Answer

The value of dydx\frac{dy}{dx} at x=2x = 2 is −75-75.
Applying the Quotient Rule u′v−uv′v2\frac{u'v - uv'}{v^2} along with the Chain Rule to differentiate u(x)=(x2+1)3u(x) = (x^2+1)^3 yields u′(x)=6x(x2+1)2u'(x) = 6x(x^2+1)^2. Evaluating at x=2x=2 gives u(2)=125u(2)=125, u′(2)=300u'(2)=300, v(2)=1v(2)=1, and v′(2)=3v'(2)=3, leading to 300(1)−125(3)12=−75\frac{300(1) - 125(3)}{1^2} = -75.

Step-by-Step Solution

1
Set up the Quotient Rule framework
Let u(x)=(x2+1)3u(x) = (x^2 + 1)^3 and v(x)=3x−5v(x) = 3x - 5, so that y=u(x)v(x)y = \frac{u(x)}{v(x)} and dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
The function is expressed as a quotient of two differentiable terms.
2
Differentiate the numerator using the Chain Rule
u′(x)=3(x2+1)2⋅ddx(x2+1)=6x(x2+1)2u'(x) = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 6x(x^2 + 1)^2.
The numerator is a composite function requiring the inner derivative derivative of x2+1x^2+1 to be multiplied.
3
Differentiate the denominator
v′(x)=3v'(x) = 3.
The derivative of a linear function 3x−53x - 5 with respect to xx is its coefficient 3.
4
Evaluate u(x)u(x), u′(x)u'(x), v(x)v(x), and v′(x)v'(x) at x=2x = 2
u(2)=125u(2) = 125, u′(2)=300u'(2) = 300, v(2)=1v(2) = 1, v′(2)=3v'(2) = 3.
Substituting x=2x = 2 into each evaluated component simplifies the numerical calculation.
5
Substitute numerical values into the Quotient Rule formula
dydx∣x=2=(300)(1)−(125)(3)12=300−3751=−75\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(3)}{1^2} = \frac{300 - 375}{1} = -75.
Completing the arithmetic calculation yields the final numerical derivative value.

Key Concept

Combining the Quotient Rule and Chain Rule for composite fractional functions
Estimated Time:1m 30s
Question 2Question →

Given the function y=x23x2+4y = x^2 \sqrt{3x^2 + 4}, find the numerical value of dydx\frac{dy}{dx} at x=2x = 2.

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Answer: 22

Answer

The numerical value of the derivative at x=2x = 2 is 22.
Applying both the Product Rule and Chain Rule correctly yields dydx=2x3x2+4+3x33x2+4\frac{dy}{dx} = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}. Evaluating this expression at x=2x = 2 gives 16+6=2216 + 6 = 22.

Step-by-Step Solution

1
Decompose the function into a product of two functions
Let u(x)=x2u(x) = x^2 and v(x)=3x2+4=(3x2+4)1/2v(x) = \sqrt{3x^2 + 4} = (3x^2 + 4)^{1/2}.
The function yy is expressed as the product of u(x)u(x) and v(x)v(x), requiring the Product Rule for differentiation.
2
Find the derivative of the inner square root function using the Chain Rule
v′(x)=12(3x2+4)−1/2⋅6x=3x3x2+4v'(x) = \frac{1}{2}(3x^2 + 4)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{3x^2 + 4}}.
The Chain Rule states that ddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x).
3
Combine derivatives using the Product Rule formula
dydx=u′(x)v(x)+u(x)v′(x)=2x3x2+4+3x33x2+4\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}.
The Product Rule formula is ddx[u⋅v]=u′v+uv′\frac{d}{dx}[u \cdot v] = u'v + uv'.
4
Evaluate the derivative at x=2x = 2
dydx∣x=2=2(2)3(2)2+4+3(2)33(2)2+4=4(4)+244=16+6=22\frac{dy}{dx}\Big|_{x=2} = 2(2)\sqrt{3(2)^2 + 4} + \frac{3(2)^3}{\sqrt{3(2)^2 + 4}} = 4(4) + \frac{24}{4} = 16 + 6 = 22.
Substituting x=2x = 2 gives the specific slope of the tangent line to the curve at that point.

Key Concept

Combining the Product Rule and Chain Rule to evaluate derivatives of composite product functions
Question 3Question →

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

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Answer: 54

Answer

54
Applying the chain rule gives dydx=3(2x+1)2⋅2=6(2x+1)2\frac{dy}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2. Evaluating at x=1x = 1 gives 6(3)2=546(3)^2 = 54.

Step-by-Step Solution

1
Differentiate y=(2x+1)3y = (2x + 1)^3 using the chain rule.
dydx=6(2x+1)2\frac{dy}{dx} = 6(2x + 1)^2
According to the chain rule, ddx[un]=nun−1⋅dudx\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx}, where u=2x+1u = 2x + 1 and dudx=2\frac{du}{dx} = 2.
2
Evaluate the derivative at x=1x = 1.
dydx∣x=1=54\frac{dy}{dx}\Big|_{x=1} = 54
Substituting x=1x = 1 into 6(2x+1)26(2x + 1)^2 yields 6(3)2=546(3)^2 = 54.

Key Concept

Chain Rule for Differentiation
Question 4Question →

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

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Answer: 112112

Answer

112
Applying the product rule dydx=u′v+uv′\frac{dy}{dx} = u'v + uv' along with the chain rule gives u′(1)=6u'(1) = 6, v(1)=16v(1) = 16, u(1)=1u(1) = 1, and v′(1)=16v'(1) = 16. Evaluating u′(1)v(1)+u(1)v′(1)u'(1)v(1) + u(1)v'(1) yields 6×16+1×16=96+16=1126 \times 16 + 1 \times 16 = 96 + 16 = 112.

Step-by-Step Solution

1
Decompose the function into two components for the Product Rule
Let u(x)=(2x−1)3u(x) = (2x - 1)^3 and v(x)=(x2+3)2v(x) = (x^2 + 3)^2, so y=u(x)v(x)y = u(x) v(x).
The function is a product of two composite expressions.
2
Differentiate each term using the Chain Rule
u′(x)=3(2x−1)2⋅ddx(2x−1)=6(2x−1)2u'(x) = 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 6(2x - 1)^2
v′(x)=2(x2+3)⋅ddx(x2+3)=4x(x2+3)v'(x) = 2(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) = 4x(x^2 + 3)
Applying the Chain Rule requires differentiating the outer function and multiplying by the derivative of the inner function.
3
Apply the Product Rule formula
dydx=u′(x)v(x)+u(x)v′(x)=6(2x−1)2(x2+3)2+4x(2x−1)3(x2+3)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 6(2x - 1)^2 (x^2 + 3)^2 + 4x(2x - 1)^3 (x^2 + 3)
The derivative of a product u⋅vu \cdot v is u′v+uv′u'v + uv'.
4
Evaluate the derivative at x=1x = 1
u(1)=(2(1)−1)3=1u(1) = (2(1) - 1)^3 = 1
u′(1)=6(2(1)−1)2=6u'(1) = 6(2(1) - 1)^2 = 6
v(1)=(12+3)2=16v(1) = (1^2 + 3)^2 = 16
v′(1)=4(1)(12+3)=16v'(1) = 4(1)(1^2 + 3) = 16
dydx∣x=1=(6)(16)+(1)(16)=96+16=112\left.\frac{dy}{dx}\right|_{x=1} = (6)(16) + (1)(16) = 96 + 16 = 112
Substitute x=1x = 1 into all expressions to find the numerical derivative.

Key Concept

Combined Product Rule and Chain Rule of Differentiation
Estimated Time:2m 0s
Question 5Question →

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

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Answer: 34

Answer

The value of dydx\frac{dy}{dx} at x=2x = 2 is 3434.
Applying the product rule dydx=u′v+uv′\frac{dy}{dx} = u'v + uv' alongside the chain rule gives dydx=2x(2x−3)3+6(x2+1)(2x−3)2\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2. Evaluating at x=2x = 2 yields 4(1)+30(1)=344(1) + 30(1) = 34.

Step-by-Step Solution

1
Identify component functions for the product rule
Let u(x)=x2+1u(x) = x^2 + 1 and v(x)=(2x−3)3v(x) = (2x - 3)^3.
The function yy is a product of two differentiable functions.
2
Differentiate each component function
u′(x)=2xu'(x) = 2x and v′(x)=3(2x−3)2⋅2=6(2x−3)2v'(x) = 3(2x - 3)^2 \cdot 2 = 6(2x - 3)^2.
The power rule gives u′(x)u'(x) and the chain rule gives v′(x)v'(x) by multiplying by the derivative of the inner function (2x−3)(2x - 3).
3
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)
dydx=2x(2x−3)3+6(x2+1)(2x−3)2\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2.
To find the general derivative of a product of functions.
4
Evaluate the derivative at x=2x = 2
dydx∣x=2=2(2)(2(2)−3)3+6(22+1)(2(2)−3)2=4(1)+30(1)=34\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34.
To calculate the specific numerical value of the derivative at x=2x = 2.

Key Concept

Product and Chain Rules of Differentiation
Estimated Time:1m 30s
Question 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

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 11Question →

If y=(2x−3)5y = (2x - 3)^5, what is the value of dydx\frac{dy}{dx} at x=2x = 2?

Show answer & explanation

Answer: 10

Answer

10
The derivative of y=(2x−3)5y = (2x - 3)^5 with respect to xx requires the chain rule: dydx=5(2x−3)4⋅ddx(2x−3)=5(2x−3)4⋅2=10(2x−3)4\frac{dy}{dx} = 5(2x - 3)^4 \cdot \frac{d}{dx}(2x - 3) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4. Substituting x=2x = 2 gives 10(2(2)−3)4=10(1)4=1010(2(2) - 3)^4 = 10(1)^4 = 10, which makes 1010 the correct value.

Step-by-Step Solution

1
Identify the inner function u(x)u(x) and outer function f(u)f(u)
Let u=2x−3u = 2x - 3, so y=u5y = u^5.
The chain rule states that dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
2
Differentiate yy with respect to uu and uu with respect to xx
dydu=5u4=5(2x−3)4\frac{dy}{du} = 5u^4 = 5(2x - 3)^4 and dudx=2\frac{du}{dx} = 2.
Apply the power rule to both functions.
3
Multiply the derivatives to find dydx\frac{dy}{dx}
dydx=5(2x−3)4⋅2=10(2x−3)4\frac{dy}{dx} = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4.
Combine terms using the chain rule formula.
4
Substitute x=2x = 2 into the derivative
dydx∣x=2=10(2(2)−3)4=10(1)4=10\frac{dy}{dx}\Big|_{x=2} = 10(2(2) - 3)^4 = 10(1)^4 = 10.
Evaluate the expression numerically at the target point.

Key Concept

Chain Rule for Differentiation
Estimated Time:45s
Question 12Question →

If y=x3(2x−1)4y = x^3(2x - 1)^4, find the value of dydx\frac{dy}{dx} at x=1x = 1.

Show answer & explanation

Answer: 11

Answer

The numerical value of dydx\frac{dy}{dx} at x=1x = 1 is 1111.
Applying the product rule dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} along with the chain rule for the expression (2x−1)4(2x - 1)^4 gives dydx=8x3(2x−1)3+3x2(2x−1)4\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4. Evaluating this at x=1x = 1 gives 8(1)3(1)3+3(1)2(1)4=8+3=118(1)^3(1)^3 + 3(1)^2(1)^4 = 8 + 3 = 11.

Step-by-Step Solution

1
Set up the product rule for y=u⋅vy = u \cdot v, where u=x3u = x^3 and v=(2x−1)4v = (2x - 1)^4.
u=x3u = x^3 and v=(2x−1)4v = (2x - 1)^4
The function is expressed as the product of two algebraic terms.
2
Find the derivative of each function component.
dudx=3x2\frac{du}{dx} = 3x^2 and dvdx=8(2x−1)3\frac{dv}{dx} = 8(2x - 1)^3
The power rule gives dudx=3x2\frac{du}{dx} = 3x^2, and applying the chain rule to (2x−1)4(2x - 1)^4 yields 4(2x−1)3⋅2=8(2x−1)34(2x - 1)^3 \cdot 2 = 8(2x - 1)^3.
3
Substitute components into the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.
dydx=8x3(2x−1)3+3x2(2x−1)4\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4
Combining udvdxu\frac{dv}{dx} and vdudxv\frac{du}{dx} provides the full expression for the derivative.
4
Evaluate the derivative at x=1x = 1.
dydx∣x=1=8(1)3(2(1)−1)3+3(1)2(2(1)−1)4=8+3=11\frac{dy}{dx}\Big|_{x=1} = 8(1)^3(2(1) - 1)^3 + 3(1)^2(2(1) - 1)^4 = 8 + 3 = 11
Substituting x=1x = 1 simplifies the terms to 8(1)+3(1)=118(1) + 3(1) = 11.

Key Concept

Rules of Differentiation (Product and Chain Rules)
Question 13Question →

If y=4x+1x−2y = \frac{4x + 1}{x - 2}, calculate the numerical value of dydx\frac{dy}{dx} at x=3x = 3.

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Answer: -9

Answer

The numerical value of dydx\frac{dy}{dx} at x=3x = 3 is −9-9.
Using the quotient rule dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} for y=4x+1x−2y = \frac{4x + 1}{x - 2} gives dydx=(x−2)(4)−(4x+1)(1)(x−2)2=−9(x−2)2\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{-9}{(x - 2)^2}. Substituting x=3x = 3 produces −9(3−2)2=−9\frac{-9}{(3 - 2)^2} = -9.

Step-by-Step Solution

1
Identify u(x)u(x) and v(x)v(x) for the quotient rule formula.
u=4x+1u = 4x + 1 and v=x−2v = x - 2.
The given equation y=4x+1x−2y = \frac{4x + 1}{x - 2} is a quotient of two functions of xx.
2
Calculate the individual derivatives dudx\frac{du}{dx} and dvdx\frac{dv}{dx}.
dudx=4\frac{du}{dx} = 4 and dvdx=1\frac{dv}{dx} = 1.
These derivatives are required components of the quotient rule.
3
Substitute the expressions into the quotient rule formula dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2} and simplify.
dydx=(x−2)(4)−(4x+1)(1)(x−2)2=4x−8−4x−1(x−2)2=−9(x−2)2\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{4x - 8 - 4x - 1}{(x - 2)^2} = \frac{-9}{(x - 2)^2}.
Simplifying the numerator yields the general derivative function.
4
Evaluate the derivative at x=3x = 3.
dydx∣x=3=−9(3−2)2=−91=−9\frac{dy}{dx}\Big|_{x=3} = \frac{-9}{(3 - 2)^2} = \frac{-9}{1} = -9.
Substituting x=3x = 3 gives the requested numerical value.

Key Concept

Quotient Rule of Differentiation
Question 14Question →

What is the derivative of the function f(x)=sin⁡(2x)(1+x2)2f(x) = \frac{\sin(2x)}{(1 + x^2)^2} with respect to xx?

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Answer: 2[(1+x2)cos⁡(2x)−2xsin⁡(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Answer

2[(1+x2)cos⁡(2x)−2xsin⁡(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}
Applying the quotient rule u′v−uv′v2\frac{u'v - uv'}{v^2} with u=sin⁡(2x)u = \sin(2x) and v=(1+x2)2v = (1 + x^2)^2 yields u′=2cos⁡(2x)u' = 2\cos(2x) and v′=4x(1+x2)v' = 4x(1 + x^2). Substituting these into the formula gives 2cos⁡(2x)(1+x2)2−4xsin⁡(2x)(1+x2)(1+x2)4\frac{2\cos(2x)(1+x^2)^2 - 4x\sin(2x)(1+x^2)}{(1+x^2)^4}. Factoring out (1+x2)(1+x^2) from numerator and denominator reduces the expression to 2[(1+x2)cos⁡(2x)−2xsin⁡(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}.

Step-by-Step Solution

1
Identify the numerator u(x)u(x) and denominator v(x)v(x) for the quotient rule.
u(x)=sin⁡(2x)u(x) = \sin(2x) and v(x)=(1+x2)2v(x) = (1 + x^2)^2
The quotient rule states that ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.
2
Find the derivatives u′(x)u'(x) and v′(x)v'(x) using the chain rule.
u′(x)=2cos⁡(2x)u'(x) = 2\cos(2x) and v′(x)=2(1+x2)⋅(2x)=4x(1+x2)v'(x) = 2(1 + x^2) \cdot (2x) = 4x(1 + x^2)
By the chain rule, ddxsin⁡(2x)=cos⁡(2x)⋅2\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2 and ddx(1+x2)2=2(1+x2)⋅2x\frac{d}{dx}(1 + x^2)^2 = 2(1 + x^2) \cdot 2x.
3
Substitute u,u′,v,v′u, u', v, v' into the quotient rule formula.
f′(x)=2cos⁡(2x)(1+x2)2−sin⁡(2x)⋅4x(1+x2)((1+x2)2)2f'(x) = \frac{2\cos(2x)(1 + x^2)^2 - \sin(2x) \cdot 4x(1 + x^2)}{((1 + x^2)^2)^2}
Applying u′v−uv′v2\frac{u'v - uv'}{v^2} directly.
4
Factor out (1+x2)(1 + x^2) from the numerator and simplify the denominator.
f′(x)=(1+x2)[2cos⁡(2x)(1+x2)−4xsin⁡(2x)](1+x2)4=2(1+x2)cos⁡(2x)−4xsin⁡(2x)(1+x2)3f'(x) = \frac{(1 + x^2)[2\cos(2x)(1 + x^2) - 4x\sin(2x)]}{(1 + x^2)^4} = \frac{2(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}
Canceling the common factor (1+x2)(1 + x^2) from numerator and denominator.
5
Factor out the constant 22 from the numerator.
f′(x)=2[(1+x2)cos⁡(2x)−2xsin⁡(2x)](1+x2)3f'(x) = \frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}
Putting the derivative into simplified factored form.

Key Concept

Combining Quotient Rule and Chain Rule for Composite Functions
Question 15Question →

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

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Answer: 24

Answer

24
Applying the chain rule gives dydx=3(x2−2x+2)2(2x−2)\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2). Evaluating this derivative at x=2x = 2 yields 3(2)2(2)=243(2)^2(2) = 24.

Step-by-Step Solution

1
Identify the composite structure and state the chain rule formula.
Let u=x2−2x+2u = x^2 - 2x + 2, so y=u3y = u^3. By the chain rule, dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
The function is an expression raised to a power, requiring the chain rule.
2
Differentiate the outer function with respect to uu and the inner function with respect to xx.
\frac{dy}{du} = 3(x^2 - 2x + 2)^2 and \frac{du}{dx} = 2x - 2.
Apply the power rule to both the outer expression and inner polynomial.
3
Multiply the two derivatives to express dydx\frac{dy}{dx}.
\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2).
Product of outer and inner derivatives per the chain rule.
4
Evaluate the expression at x=2x = 2.
\frac{dy}{dx}\Big|_{x=2} = 3(4 - 4 + 2)^2(4 - 2) = 3(2)^2(2) = 24.
Substitute x=2x = 2 into the derivative expression and simplify.

Key Concept

Chain Rule of Differentiation
Question 16Question →

If y=(2x2+1)34x−1y = \frac{(2x^2 + 1)^3}{4x - 1}, determine the numerical value of dydx\frac{dy}{dx} at x=1x = 1.

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Answer: 24

Answer

The numerical value of the derivative dydx\frac{dy}{dx} at x=1x = 1 is 24.
Evaluating the derivative requires combined application of the quotient rule and chain rule. Differentiating u(x)=(2x2+1)3u(x) = (2x^2 + 1)^3 via the chain rule yields u′(x)=12x(2x2+1)2u'(x) = 12x(2x^2 + 1)^2. At x=1x = 1, u(1)=27u(1) = 27, u′(1)=108u'(1) = 108, v(1)=3v(1) = 3, and v′(1)=4v'(1) = 4. Substituting these into the quotient rule u′v−uv′v2\frac{u'v - uv'}{v^2} produces 324−1089=24\frac{324 - 108}{9} = 24.

Step-by-Step Solution

1
Decompose the function for the Quotient Rule
Let u(x)=(2x2+1)3u(x) = (2x^2 + 1)^3 and v(x)=4x−1v(x) = 4x - 1.
The given expression is a quotient of two differentiable functions.
2
Differentiate the numerator using the Chain Rule
u′(x)=3(2x2+1)2⋅4x=12x(2x2+1)2u'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2
The chain rule states that ddx[g(x)]n=n[g(x)]n−1⋅g′(x)\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x).
3
Differentiate the denominator and evaluate all components at x=1x = 1
v′(1)=4v'(1) = 4, v(1)=3v(1) = 3, u(1)=27u(1) = 27, and u′(1)=108u'(1) = 108
Evaluating each term individually reduces algebraic complexity prior to applying the quotient rule.
4
Apply the Quotient Rule formula and evaluate
dydx∣x=1=(108)(3)−(27)(4)32=324−1089=24\frac{dy}{dx}\Big|_{x=1} = \frac{(108)(3) - (27)(4)}{3^2} = \frac{324 - 108}{9} = 24
The quotient rule formula is dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.

Key Concept

Combination of Quotient Rule and Chain Rule
Estimated Time:1m 30s
Question 17Question →

If y=(3x−1)4y = (3x - 1)^4, what is the numerical value of dydx\frac{dy}{dx} evaluated at x=1x = 1?

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Answer: 96

Answer

96
Applying the chain rule ddx[un]=nun−1⋅dudx\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx} to y=(3x−1)4y = (3x - 1)^4 gives dydx=4(3x−1)3⋅3=12(3x−1)3\frac{dy}{dx} = 4(3x - 1)^3 \cdot 3 = 12(3x - 1)^3. Evaluating this expression at x=1x = 1 yields 12(3(1)−1)3=12(2)3=9612(3(1) - 1)^3 = 12(2)^3 = 96.

Step-by-Step Solution

1
Differentiate y=(3x−1)4y = (3x - 1)^4 with respect to xx using the chain rule.
dydx=12(3x−1)3\frac{dy}{dx} = 12(3x - 1)^3
Differentiate the outer power function to get 4(3x−1)34(3x - 1)^3 and multiply by the derivative of the inner linear function 3x−13x - 1, which is 3.
2
Substitute x=1x = 1 into the derivative function.
96
Evaluating 12(3(1)−1)312(3(1) - 1)^3 gives 12(2)3=12×8=9612(2)^3 = 12 \times 8 = 96.

Key Concept

Chain Rule of Differentiation
Question 18Question →

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

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Answer: 2

Answer

The numerical value of the derivative at x=1x = 1 is 2.
To find dydx\frac{dy}{dx} at x=1x = 1 for y=(x+1)3x2+1y = \frac{(x + 1)^3}{x^2 + 1}, we use the quotient rule dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. Setting u=(x+1)3u = (x + 1)^3 gives dudx=3(x+1)2\frac{du}{dx} = 3(x + 1)^2, and setting v=x2+1v = x^2 + 1 gives dvdx=2x\frac{dv}{dx} = 2x. At x=1x = 1, u=8u = 8, dudx=12\frac{du}{dx} = 12, v=2v = 2, and dvdx=2\frac{dv}{dx} = 2. Substituting into the quotient formula gives (2)(12)−(8)(2)22=24−164=2\frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2.

Step-by-Step Solution

1
Identify the components of the quotient rule
Let u(x)=(x+1)3u(x) = (x + 1)^3 and v(x)=x2+1v(x) = x^2 + 1.
The given function is a quotient of two functions of xx.
2
Differentiate the numerator using the chain rule and the denominator using standard rules
u′(x)=3(x+1)2u'(x) = 3(x + 1)^2 and v′(x)=2xv'(x) = 2x.
The chain rule states that ddx[g(x)n]=n⋅g(x)n−1⋅g′(x)\frac{d}{dx}[g(x)^n] = n \cdot g(x)^{n-1} \cdot g'(x).
3
Evaluate all function components at x=1x = 1
u(1)=8u(1) = 8, u′(1)=12u'(1) = 12, v(1)=2v(1) = 2, and v′(1)=2v'(1) = 2.
Substituting x=1x = 1 simplifies the calculation before applying the full quotient expression.
4
Apply the quotient rule formula to calculate the final derivative value
dydx∣x=1=v(1)u′(1)−u(1)v′(1)[v(1)]2=(2)(12)−(8)(2)22=24−164=2\frac{dy}{dx}\Big|_{x=1} = \frac{v(1)u'(1) - u(1)v'(1)}{[v(1)]^2} = \frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2.
Substituting the numerical values yields the final result.

Key Concept

Combining the Quotient Rule and Chain Rule for differentiation
Question 19Question →

If y=x23x+1y = x^2 \sqrt{3x + 1}, what is the value of dydx\frac{dy}{dx} evaluated at x=1x = 1?

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Answer: 194\frac{19}{4}

Answer

The numerical value of dydx\frac{dy}{dx} at x=1x = 1 is 194\frac{19}{4}.
Applying the product rule to y=x23x+1y = x^2 \sqrt{3x + 1} gives dydx=x2⋅ddx(3x+1)+3x+1⋅ddx(x2)\frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\sqrt{3x + 1}) + \sqrt{3x + 1} \cdot \frac{d}{dx}(x^2). By the chain rule, ddx(3x+1)=323x+1\frac{d}{dx}(\sqrt{3x + 1}) = \frac{3}{2\sqrt{3x + 1}}. Substituting x=1x = 1 into dydx=3x223x+1+2x3x+1\frac{dy}{dx} = \frac{3x^2}{2\sqrt{3x + 1}} + 2x\sqrt{3x + 1} gives 34+4=194\frac{3}{4} + 4 = \frac{19}{4}. Thus, the option equal to 194\frac{19}{4} is correct.

Step-by-Step Solution

1
Identify the main differentiation rule required and split the function into two components.
Let u=x2u = x^2 and v=(3x+1)1/2v = (3x + 1)^{1/2}, so y=u⋅vy = u \cdot v.
The function is a product of x2x^2 and a composite function 3x+1\sqrt{3x + 1}.
2
Differentiate uu with respect to xx, and use the chain rule to differentiate vv with respect to xx.
dudx=2x\frac{du}{dx} = 2x and dvdx=12(3x+1)−1/2⋅3=323x+1\frac{dv}{dx} = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}}.
The chain rule requires multiplying the derivative of the outer square root function by the derivative of the inner linear function 3x+13x + 1.
3
Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.
dydx=x2(323x+1)+3x+1(2x)\frac{dy}{dx} = x^2 \left( \frac{3}{2\sqrt{3x + 1}} \right) + \sqrt{3x + 1} (2x).
The derivative of a product uvuv is uv′+vu′u v' + v u'.
4
Substitute x=1x = 1 into the derivative expression.
dydx∣x=1=(1)2(323(1)+1)+3(1)+1⋅2(1)=1⋅(34)+2⋅2=34+4=194\frac{dy}{dx}\Big|_{x=1} = (1)^2 \left( \frac{3}{2\sqrt{3(1) + 1}} \right) + \sqrt{3(1) + 1} \cdot 2(1) = 1 \cdot \left( \frac{3}{4} \right) + 2 \cdot 2 = \frac{3}{4} + 4 = \frac{19}{4}.
Evaluating at x=1x = 1 requires simplifying 4=2\sqrt{4} = 2 and combining fractions.

Key Concept

Combining Product Rule and Chain Rule for Differentiation
Question 20Question →

If y=x3sin⁡(2x)y = x^3 \sin(2x), what is dydx\frac{dy}{dx}?

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Answer: 3x2sin⁡(2x)+2x3cos⁡(2x)3x^2 \sin(2x) + 2x^3 \cos(2x)

Answer

The derivative dydx\frac{dy}{dx} is 3x2sin⁡(2x)+2x3cos⁡(2x)3x^2 \sin(2x) + 2x^3 \cos(2x).
Applying the product rule ddx[uv]=udvdx+vdudx\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} with u=x3u = x^3 and v=sin⁡(2x)v = \sin(2x) yields x3(2cos⁡(2x))+sin⁡(2x)(3x2)x^3(2\cos(2x)) + \sin(2x)(3x^2), which simplifies to 3x2sin⁡(2x)+2x3cos⁡(2x)3x^2 \sin(2x) + 2x^3 \cos(2x).

Step-by-Step Solution

1
Identify the two component functions for the Product Rule
Let u=x3u = x^3 and v=sin⁡(2x)v = \sin(2x), so y=uvy = uv.
The function is a product of an algebraic term and a trigonometric term.
2
Find the derivative of each component function
\frac{du}{dx} = 3x^2 and \frac{dv}{dx} = 2\cos(2x) using the chain rule for \sin(2x).
The derivative of \sin(2x) with respect to 2x is \cos(2x), multiplied by the derivative of 2x which is 2.
3
Apply the Product Rule formula \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
\frac{dy}{dx} = x^3 \cdot 2\cos(2x) + \sin(2x) \cdot 3x^2 = 3x^2 \sin(2x) + 2x^3 \cos(2x).
Combining the differentiated terms according to the product rule gives the final derivative.

Key Concept

Product Rule combined with Chain Rule for trigonometric functions
Estimated Time:1m 30s
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Rules of Differentiation (Product, Quotient, and Chain Rules) Practice Questions — JAMB UTME | Examkin