Calculus Alıştırma Soruları — JAMB UTME — Sayfa 7

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

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Cevap: 2

Cevap

The numerical value of the derivative at

x = 1

is 2.

To find

\frac{dy}{dx}

at

x = 1

for

y = \frac{(x + 1)^3}{x^2 + 1}

, we use the quotient rule

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

. Setting

u = (x + 1)^3

gives

\frac{du}{dx} = 3(x + 1)^2

, and setting

v = x^2 + 1

gives

\frac{dv}{dx} = 2x

. At

x = 1

,

u = 8

,

\frac{du}{dx} = 12

,

v = 2

, and

\frac{dv}{dx} = 2

. Substituting into the quotient formula gives

\frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

.

Adım Adım Çözüm

1

Identify the components of the quotient rule

Let

u(x) = (x + 1)^3

and

v(x) = x^2 + 1

.

The given function is a quotient of two functions of

x

.

2

Differentiate the numerator using the chain rule and the denominator using standard rules

u'(x) = 3(x + 1)^2

and

v'(x) = 2x

.

The chain rule states that

\frac{d}{dx}[g(x)^n] = n \cdot g(x)^{n-1} \cdot g'(x)

.

3

Evaluate all function components at

x = 1

u(1) = 8

,

u'(1) = 12

,

v(1) = 2

, and

v'(1) = 2

.

Substituting

x = 1

simplifies the calculation before applying the full quotient expression.

4

Apply the quotient rule formula to calculate the final derivative value

\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.

Anahtar Kavram

Combining the Quotient Rule and Chain Rule for differentiation

Soru 122Soru →

A curve is defined by the equation $y = 4x + \frac{9}{x}$ for $x > 0$ . What is the $y$ -value at the minimum stationary point of the curve?

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Cevap: 12

Cevap

The

y

-value at the minimum stationary point is 12.

To find the minimum value of

y = 4x + \frac{9}{x}

for

x > 0

, set the first derivative

\frac{dy}{dx} = 4 - \frac{9}{x^2}

equal to

0

, yielding

x = 1.5

. The second derivative

\frac{d^2y}{dx^2} = \frac{18}{x^3}

is positive at

x = 1.5

, confirming a minimum stationary point. Evaluating the original equation at

x = 1.5

gives

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

.

Adım Adım Çözüm

1

Differentiate the given function

y = 4x + 9x^{-1}

with respect to

x

.

\frac{dy}{dx} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}

Stationary points occur where the gradient (first derivative) equals zero.

2

Set the first derivative to zero and solve for

x

given the domain constraint

x > 0

.

4 - \frac{9}{x^2} = 0 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2} = 1.5

Solving

\frac{dy}{dx} = 0

yields the

x

-coordinate of the turning point.

3

Evaluate the second derivative to confirm the turning point is a local minimum.

\frac{d^2y}{dx^2} = \frac{18}{x^3}. \text{ At } x = 1.5, \frac{d^2y}{dx^2} = \frac{18}{3.375} = 5.333 > 0

A positive second derivative indicates that the stationary point is a local minimum.

4

Substitute

x = 1.5

back into the original equation

y = 4x + \frac{9}{x}

to find the corresponding

y

-value.

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

The question asks for the

y

-value of the curve at the minimum stationary point.

Anahtar Kavram

Stationary Points, Maxima, and Minima

Tahmini Süre:2m 0s

Soru 123Soru →

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

\frac{19}{4}

\frac{17}{4}

\frac{13}{4}

\frac{3}{2}

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Cevap: $\frac{19}{4}$

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is

\frac{19}{4}

.

Applying the product rule to

y = x^2 \sqrt{3x + 1}

gives

\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,

\frac{d}{dx}(\sqrt{3x + 1}) = \frac{3}{2\sqrt{3x + 1}}

. Substituting

x = 1

into

\frac{dy}{dx} = \frac{3x^2}{2\sqrt{3x + 1}} + 2x\sqrt{3x + 1}

gives

\frac{3}{4} + 4 = \frac{19}{4}

. Thus, the option equal to

\frac{19}{4}

is correct.

Adım Adım Çözüm

1

Identify the main differentiation rule required and split the function into two components.

Let

u = x^2

and

v = (3x + 1)^{1/2}

, so

y = u \cdot v

.

The function is a product of

x^2

and a composite function

\sqrt{3x + 1}

.

2

Differentiate

u

with respect to

x

, and use the chain rule to differentiate

v

with respect to

x

.

\frac{du}{dx} = 2x

and

\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 + 1

.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

.

\frac{dy}{dx} = x^2 \left( \frac{3}{2\sqrt{3x + 1}} \right) + \sqrt{3x + 1} (2x)

.

The derivative of a product

uv

is

u v' + v u'

.

4

Substitute

x = 1

into the derivative expression.

\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 = 1

requires simplifying

\sqrt{4} = 2

and combining fractions.

Anahtar Kavram

Combining Product Rule and Chain Rule for Differentiation

Soru 124Soru →

Find the area of the region bounded by the curve $y = 3x^2$ , the $x$ -axis, and the vertical lines $x = 1$ and $x = 3$ .

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Cevap: 26

Cevap

The area of the bounded region is 26 square units.

The area under

y = 3x^2

from

x = 1

to

x = 3

is calculated using the definite integral

\int_{1}^{3} 3x^2 \, dx = [x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

square units.

Adım Adım Çözüm

1

Set up the definite integral representing the bounded area.

A = \int_{1}^{3} 3x^2 \, dx

The area under a non-negative curve

y = f(x)

from

x = a

to

x = b

above the

x

-axis is given by the definite integral

\int_{a}^{b} f(x) \, dx

.

2

Determine the antiderivative of

3x^2

.

\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3

Applying the power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

gives

x^3

.

3

Evaluate the definite integral using the fundamental theorem of calculus.

[x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

Substitute the upper limit

x = 3

and subtract the value of the function evaluated at the lower limit

x = 1

.

Anahtar Kavram

Area under a curve using definite integration

Soru 125Soru →

Using differentiation from first principles, what is the value of the derivative of the function $f(x) = x^2 + 2x$ at the point where $x = 3$ ?

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Cevap: 8

Cevap

The derivative of

f(x) = x^2 + 2x

evaluated at

x = 3

is 8.

Applying first principles to

f(x) = x^2 + 2x

yields

f'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - (x^2 + 2x)}{h} = \lim_{h \to 0} (2x + h + 2) = 2x + 2

. Evaluating this derivative at

x = 3

gives

2(3) + 2 = 8

.

Adım Adım Çözüm

1

Apply the definition of differentiation from first principles

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

First principles uses the limit of the difference quotient to compute the instantaneous rate of change.

2

Substitute

f(x) = x^2 + 2x

into the difference quotient

\frac{[(x+h)^2 + 2(x+h)] - (x^2 + 2x)}{h} = \frac{2xh + h^2 + 2h}{h}

Expanding terms allows cancellation of non-

h

terms in the numerator.

3

Simplify the fraction by dividing by

h

2x + h + 2

Dividing out

h

removes the indeterminate form

\frac{0}{0}

.

4

Evaluate the limit as

h \to 0

and substitute

x = 3

f'(3) = 2(3) + 2 = 8

Setting

h = 0

yields the derivative function

f'(x) = 2x + 2

, which evaluates to 8 at

x = 3

.

Anahtar Kavram

Differentiation from First Principles

Soru 126Soru →

The line $3x - y + k = 0$ is a normal to the curve $y = \frac{x + 2}{x - 1}$ at a point $P$ located in the first quadrant. What is the value of the constant $k$ ?

-10

-2

10

6

Cevabı ve açıklamayı göster

Cevap: -10

Cevap

The value of the constant

k

is

-10

.

Differentiating

y = \frac{x + 2}{x - 1}

yields

\frac{dy}{dx} = \frac{-3}{(x - 1)^2}

. The gradient of the normal line is

m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}

. Equating

m_N

to the gradient of

3x - y + k = 0

(which is

3

) gives

\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9

. Solving gives

x = 4

or

x = -2

. Since point

P

lies in the first quadrant,

x = 4

, which gives

y = \frac{4 + 2}{4 - 1} = 2

. Substituting

(4, 2)

into

3x - y + k = 0

gives

3(4) - 2 + k = 0

, so

k = -10

.

Adım Adım Çözüm

1

Differentiate the curve equation to find the tangent gradient function.

Using the quotient rule on

y = \frac{x + 2}{x - 1}

,

\frac{dy}{dx} = \frac{1(x - 1) - 1(x + 2)}{(x - 1)^2} = \frac{-3}{(x - 1)^2}

.

The derivative gives the gradient of the tangent line to the curve at any point

x

.

2

Determine the gradient function for the normal line.

The normal gradient is

m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}

.

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of

\frac{dy}{dx}

.

3

Equate the normal gradient to the gradient of the given line and solve for

x

.

The line

3x - y + k = 0

has gradient

3

. Setting

\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9 \implies x - 1 = \pm 3

, giving

x = 4

or

x = -2

.

The normal line at point

P

must be parallel to (and thus have the same gradient as) the given line

3x - y + k = 0

.

4

Select the coordinate corresponding to the first quadrant and calculate the

y

-coordinate.

Since

P

lies in the first quadrant (

x > 0, y > 0

),

x = 4

. Substituting

x = 4

into the curve equation gives

y = \frac{4 + 2}{4 - 1} = 2

. Thus,

P = (4, 2)

.

The problem specifies that point

P

is located in the first quadrant.

5

Substitute point

P(4, 2)

into the line equation to solve for

k

.

Substituting

x = 4

and

y = 2

into

3x - y + k = 0

yields

3(4) - 2 + k = 0 \implies 10 + k = 0 \implies k = -10

.

Point

P

lies on the normal line, so its coordinates must satisfy the line equation.

Anahtar Kavram

The normal line to a curve at a point

P

is perpendicular to the tangent line at

P

, with gradient

m_{\text{normal}} = -\frac{1}{dy/dx}

. A point

P

on the curve must satisfy both the curve equation and the normal line equation.

Soru 127Soru →

What is the value of the definite integral $\int_{0}^{4} (x + 1) \, dx$ ?

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Cevap: 12

Cevap

The value of the definite integral is 12.

Integrating

x + 1

yields

\frac{x^2}{2} + x

. Substituting the upper limit

x = 4

gives

\frac{4^2}{2} + 4 = 8 + 4 = 12

. Substituting the lower limit

x = 0

gives

0

. Calculating

12 - 0

gives the final answer of 12.

Adım Adım Çözüm

1

Find the antiderivative of the integrand

f(x) = x + 1

.

\int (x + 1) \, dx = \frac{x^2}{2} + x + C

Apply the power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

to

x^1

and

1 = x^0

.

2

Evaluate the antiderivative at the upper bound

x = 4

and lower bound

x = 0

.

F(4) = \frac{4^2}{2} + 4 = \frac{16}{2} + 4 = 12

, and

F(0) = \frac{0^2}{2} + 0 = 0

Substitute the bounds into the antiderivative expression

F(x) = \frac{x^2}{2} + x

.

3

Apply the Fundamental Theorem of Calculus

F(b) - F(a)

.

F(4) - F(0) = 12 - 0 = 12

Subtract the lower bound result from the upper bound result.

Anahtar Kavram

Evaluation of definite integrals of simple linear polynomials

Alternatif Yöntem

The integral

\int_{0}^{4} (x + 1) \, dx

represents the area of a trapezoid under

y = x + 1

from

x = 0

to

x = 4

. The parallel vertical sides are

h_1 = 1

and

h_2 = 5

, with width

b = 4

. Using the area formula for a trapezoid:

\text{Area} = \frac{1+5}{2} \times 4 = 3 \times 4 = 12

.

Tahmini Süre:45s

Soru 128Soru →

What is the value of the definite integral $\int_{0}^{\frac{\pi}{6}} \cos(3x) \, dx$ ?

\frac{1}{3}

1

-\frac{1}{3}

\frac{1}{6}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{3}$

Cevap

The value of the definite integral is

\frac{1}{3}

.

The antiderivative of

\cos(3x)

is

\frac{1}{3}\sin(3x)

. Substituting the upper boundary

x = \frac{\pi}{6}

yields

\frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}

, and substituting the lower boundary

x = 0

yields

0

. Subtracting the lower bound evaluation from the upper bound evaluation gives

\frac{1}{3}

.

Adım Adım Çözüm

1

Find the indefinite integral of

\cos(3x)

\int \cos(3x) \, dx = \frac{1}{3}\sin(3x) + C

Using the standard integration rule

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C

.

2

Apply the upper limit of integration

x = \frac{\pi}{6}

\frac{1}{3}\sin\left(3 \cdot \frac{\pi}{6}\right) = \frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}(1) = \frac{1}{3}

Substituting the upper limit into the antiderivative.

3

Apply the lower limit of integration

x = 0

and subtract

\frac{1}{3} - \frac{1}{3}\sin(0) = \frac{1}{3} - 0 = \frac{1}{3}

Evaluating the antiderivative at the limits according to the Fundamental Theorem of Calculus.

Anahtar Kavram

Definite Integration of Trigonometric Functions

Tahmini Süre:1m 30s

Soru 129Soru →

What is the numerical value of the limit

\lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}{x}

?

\frac{1}{2}

0

2

1

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{2}$

Cevap

The numerical value of the limit is $\frac{1}{2}$.

Rationalizing the numerator by multiplying with its conjugate gives $\frac{2x}{x(\sqrt{4+x} + \sqrt{4-x})}$. Canceling $x$ yields $\frac{2}{\sqrt{4+x} + \sqrt{4-x}}$, which evaluates to $\frac{2}{4} = \frac{1}{2}$ as $x \to 0$.

Adım Adım Çözüm

1

Identify the indeterminate form

Substituting $x = 0$ directly yields $\frac{\sqrt{4} - \sqrt{4}}{0} = \frac{0}{0}$, which is an indeterminate form requiring rationalization.

Direct evaluation cannot give the true limit value when an indeterminate form is encountered.

2

Rationalize the numerator

Multiply the numerator and denominator by the conjugate $\sqrt{4 + x} + \sqrt{4 - x}$:

\frac{(\sqrt{4 + x} - \sqrt{4 - x})(\sqrt{4 + x} + \sqrt{4 - x})}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{(4 + x) - (4 - x)}{x(\sqrt{4 + x} + \sqrt{4 - x})}

The difference of squares identity $(a - b)(a + b) = a^2 - b^2$ eliminates the radical signs in the numerator.

3

Simplify the numerator and cancel common terms

$\frac{4 + x - 4 + x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2}{\sqrt{4 + x} + \sqrt{4 - x}}$

Canceling the common factor $x$ eliminates the zero-causing term in the denominator.

4

Evaluate the simplified limit as $x \to 0$

$\frac{2}{\sqrt{4 + 0} + \sqrt{4 - 0}} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$

Direct substitution is now valid since the expression is continuous at $x = 0$.

Anahtar Kavram

Evaluation of algebraic limits of indeterminate form 0/0 using radical rationalization.

Soru 130Soru →

If $y = \int (6x - 5\sin(x)) \, dx$ and $y = 12$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

7

Integrating

6x - 5\sin(x)

yields

3x^2 + 5\cos(x) + C

. Substituting

x = 0

and

y = 12

gives

12 = 3(0)^2 + 5(1) + C = 5 + C

, which gives

C = 7

.

Adım Adım Çözüm

1

Integrate each term of

6x - 5\sin(x)

with respect to

x

y = 3x^2 + 5\cos(x) + C

The integral of

6x

is

3x^2

, and the integral of

-5\sin(x)

is

+5\cos(x)

.

2

Substitute the initial conditions

x = 0

and

y = 12

to calculate

C

12 = 3(0)^2 + 5\cos(0) + C \implies 12 = 5 + C \implies C = 7

Since

\cos(0) = 1

, the right-hand side reduces to

5 + C

.

Anahtar Kavram

Indefinite Integration and Initial Value Evaluation

Soru 131Soru →

A normal line is drawn to the curve $y = \frac{x^2 + 3}{x - 1}$ at the point where $x = 2$ . Calculate the $x$ -intercept of this normal line.

Cevabı ve açıklamayı göster

Cevap: -19

Cevap

The x-intercept of the normal line is -19.

Substituting

x = 2

into the curve function gives

y = 7

. Differentiating via the quotient rule yields

\frac{dy}{dx} = \frac{x^2 - 2x - 3}{(x - 1)^2}

, which equals

-3

at

x = 2

. The perpendicular slope of the normal line is

\frac{1}{3}

. Constructing the line equation through

(2, 7)

gives

x - 3y + 19 = 0

. Setting

y = 0

produces

x = -19

.

Adım Adım Çözüm

1

Find the point of contact by substituting

x = 2

into the curve equation

y = \frac{2^2 + 3}{2 - 1} = \frac{7}{1} = 7

, so the point is

(2, 7)

The line is drawn at

x = 2

, so we need the full coordinate pair

(x_1, y_1)

2

Differentiate the curve

y = \frac{x^2 + 3}{x - 1}

using the quotient rule

\frac{dy}{dx} = \frac{(x - 1)(2x) - (x^2 + 3)(1)}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}

The derivative gives the gradient function of the curve

3

Evaluate the tangent slope

m_t

at

x = 2

m_t = \frac{2^2 - 2(2) - 3}{(2 - 1)^2} = \frac{-3}{1} = -3

Evaluating the derivative yields the gradient of the tangent at the point

4

Determine the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

Normal lines are perpendicular to tangent lines, so

m_n \cdot m_t = -1

5

Formulate the equation of the normal line passing through

(2, 7)

y - 7 = \frac{1}{3}(x - 2) \implies 3y - 21 = x - 2 \implies x - 3y + 19 = 0

Use the point-slope form

y - y_1 = m(x - x_1)

6

Find the

x

-intercept by setting

y = 0

x - 3(0) + 19 = 0 \implies x = -19

The

x

-intercept occurs where the line crosses the

x

-axis (

y = 0

)

Anahtar Kavram

Finding the equation and x-intercept of a normal line to a curve

Tahmini Süre:2m 0s

Soru 132Soru →

What is the value of the definite integral $\int_{1}^{3} (3x^2 - 2x) \, dx$ ?

12

18

20

24

Cevabı ve açıklamayı göster

Cevap: $18$

Cevap

18

Integrating

3x^2 - 2x

gives the antiderivative

F(x) = x^3 - x^2

. Evaluating

F(3) - F(1)

gives

(27 - 9) - (1 - 1) = 18 - 0 = 18

.

Adım Adım Çözüm

1

Find the indefinite integral of the function

\int (3x^2 - 2x) \, dx = x^3 - x^2

Apply the power rule of integration to each term: \int 3x^2 dx = x^3 and \int 2x dx = x^2.

2

Evaluate the antiderivative at the upper limit x = 3

(3)^3 - (3)^2 = 27 - 9 = 18

Substitute the upper limit into the antiderivative.

3

Evaluate the antiderivative at the lower limit x = 1

(1)^3 - (1)^2 = 1 - 1 = 0

Substitute the lower limit into the antiderivative.

4

Subtract the lower limit evaluation from the upper limit evaluation

18 - 0 = 18

By the Fundamental Theorem of Calculus, \int_{a}^{b} f(x) dx = F(b) - F(a).

Anahtar Kavram

Fundamental Theorem of Calculus for Definite Polynomial Integrals

Tahmini Süre:1m 0s

Soru 133Soru →

Given that $\int_{1}^{a} (3x^2 - 2x) \, dx = 48$ , where $a > 1$ is a constant, find the value of $a$ .

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the constant upper limit is 4.

Integrating

3x^2 - 2x

yields

x^3 - x^2

. Applying limits from

1

to

a

gives

(a^3 - a^2) - (1 - 1) = a^3 - a^2

. Setting

a^3 - a^2 = 48

, solving for

a > 1

gives

a = 4

because

4^3 - 4^2 = 64 - 16 = 48

.

Adım Adım Çözüm

1

Integrate the polynomial function with respect to

x

\int (3x^2 - 2x) \, dx = x^3 - x^2 + C

Using the power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

for each term.

2

Evaluate the antiderivative between the upper limit

a

and lower limit

1

[x^3 - x^2]_1^a = (a^3 - a^2) - (1^3 - 1^2) = a^3 - a^2

By the Fundamental Theorem of Calculus,

\int_{b}^{c} f(x)dx = F(c) - F(b)

.

3

Equate the expression to the given total integral value and solve for

a

a^3 - a^2 = 48 \implies a = 4

Substituting

a=4

yields

4^3 - 4^2 = 64 - 16 = 48

, which satisfies the equation.

Anahtar Kavram

Definite Integrals with Unknown Limits

Soru 134Soru →

By evaluating the limit of the difference quotient $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ , determine the value of the derivative of the function $f(x) = 2x^2 - 4x + 5$ at the point where $x = 3$ .

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Cevap: 8

Cevap

The value of the derivative of

f(x) = 2x^2 - 4x + 5

at

x = 3

is 8.

Applying first principles,

f(x+h) - f(x) = 4xh + 2h^2 - 4h

. Dividing by

h

gives

4x + 2h - 4

. Taking the limit as

h \to 0

produces

f'(x) = 4x - 4

. Substituting

x = 3

yields

4(3) - 4 = 8

.

Adım Adım Çözüm

1

Substitute

(x+h)

into the function

f(x) = 2x^2 - 4x + 5

and expand

f(x+h) = 2(x^2 + 2xh + h^2) - 4x - 4h + 5 = 2x^2 + 4xh + 2h^2 - 4x - 4h + 5

Apply algebraic expansion to determine the value of the function at

x+h

.

2

Form the difference

f(x+h) - f(x)

f(x+h) - f(x) = 4xh + 2h^2 - 4h

Subtract the original function terms to leave only terms containing

h

.

3

Divide the difference by

h

\frac{f(x+h) - f(x)}{h} = 4x + 2h - 4

Simplify the difference quotient prior to taking the limit.

4

Compute the limit as

h \to 0

f'(x) = 4x - 4

Taking the limit yields the general derivative function

f'(x)

.

5

Evaluate

f'(x)

at

x = 3

f'(3) = 4(3) - 4 = 8

Substitute

x = 3

to find the numerical rate of change at the given point.

Anahtar Kavram

Differentiation from First Principles

Soru 135Soru →

If $y = \ln(\cos(3x))$ , what is $\frac{dy}{dx}$ ?

-3\tan(3x)

3\tan(3x)

-\tan(3x)

-3\cot(3x)

Cevabı ve açıklamayı göster

Cevap: $-3\tan(3x)$

Cevap

-3\tan(3x)

Differentiating

y = \ln(\cos(3x))

requires using the chain rule twice: first for

\ln(u)

, giving

\frac{1}{\cos(3x)}

, and then for

\cos(3x)

, giving

-3\sin(3x)

. Multiplying these results yields

\frac{-3\sin(3x)}{\cos(3x)}

, which simplifies to

-3\tan(3x)

.

Adım Adım Çözüm

1

Apply the chain rule formula for differentiating logarithmic functions

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))

The outer function is

\ln(u)

where

u = \cos(3x)

.

2

Differentiate the inner trigonometric function

\cos(3x)

\frac{d}{dx}(\cos(3x)) = -3\sin(3x)

By the chain rule,

\frac{d}{dx}[\cos(kx)] = -k\sin(kx)

.

3

Multiply the terms and simplify using trigonometric identities

\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)

Using the identity

\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)

, the derivative simplifies to

-3\tan(3x)

.

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule

Tahmini Süre:1m 30s

Soru 136Soru →

Find the indefinite integral $\int \left( 3x(x - 2) + 8\cos(4x) \right) dx$ .

x^3 - 3x^2 + 2\sin(4x) + C

x^3 - 3x^2 + 2\sin(4x)

x^3 - 3x^2 - 2\sin(4x) + C

x^3 - 3x^2 + 32\sin(4x) + C

Cevabı ve açıklamayı göster

Cevap: $x^3 - 3x^2 + 2\sin(4x) + C$

Cevap

x^3 - 3x^2 + 2\sin(4x) + C

Expanding

3x(x - 2)

yields

3x^2 - 6x

. Integrating

3x^2

gives

x^3

, integrating

-6x

yields

-3x^2

, and integrating

8\cos(4x)

yields

2\sin(4x)

. Adding the arbitrary constant

C

produces

x^3 - 3x^2 + 2\sin(4x) + C

.

Adım Adım Çözüm

1

Expand the algebraic term inside the integrand.

3x(x - 2) = 3x^2 - 6x

, expressing the integral as

\int (3x^2 - 6x + 8\cos(4x)) dx

.

Expanding the product converts the term into standard polynomial form suitable for integration.

2

Integrate each component term using basic integration rules.

\int 3x^2 dx = x^3

,

\int -6x dx = -3x^2

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

.

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and trigonometric rule

\int \cos(kx) dx = \frac{\sin(kx)}{k}

.

3

Combine terms and attach the constant of integration.

x^3 - 3x^2 + 2\sin(4x) + C

An indefinite integral requires an arbitrary constant

C

to represent the full set of antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 137Soru →

Evaluate the definite integral $\int_{0}^{3} (x^2 + 2) \, dx$ .

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Cevap: 15

Cevap

The value of the definite integral is 15.

Integrating

x^2 + 2

with respect to

x

yields

\left[\frac{x^3}{3} + 2x\right]_{0}^{3}

. Substituting the upper bound

x = 3

gives

\frac{27}{3} + 6 = 15

. Substituting the lower bound

x = 0

gives

0

. The net value is

15 - 0 = 15

.

Adım Adım Çözüm

1

Find the antiderivative of

x^2 + 2

\frac{x^3}{3} + 2x

Apply the power rule of integration to each term.

2

Substitute the upper limit

x = 3

15

\frac{3^3}{3} + 2(3) = 9 + 6 = 15

3

Substitute the lower limit

x = 0

0

\frac{0^3}{3} + 2(0) = 0

4

Compute the difference between upper and lower limit values

15

15 - 0 = 15

Anahtar Kavram

Definite Integration of Polynomial Functions

Tahmini Süre:45s

Soru 138Soru →

If $y = e^{-x} \tan x + \ln(2x + 1)$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

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Cevap: 3

Cevap

The value of

\frac{dy}{dx}

at

x = 0

is 3.

Applying the product rule to

e^{-x}\tan x

gives

-e^{-x}\tan x + e^{-x}\sec^2 x

, and applying the chain rule to

\ln(2x+1)

gives

\frac{2}{2x+1}

. Evaluating

\frac{dy}{dx} = -e^{-x}\tan x + e^{-x}\sec^2 x + \frac{2}{2x+1}

at

x = 0

yields

-e^0(0) + e^0(1) + \frac{2}{1} = 3

.

Adım Adım Çözüm

1

Differentiate

u(x) = e^{-x} \tan x

using the product rule.

\frac{du}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x

The derivative of

e^{-x}

is

-e^{-x}

and the derivative of

\tan x

is \sec^2 x.

2

Differentiate

v(x) = \ln(2x + 1)

using the chain rule.

\frac{dv}{dx} = \frac{2}{2x + 1}

The derivative of

\ln(g(x))

is

\frac{g'(x)}{g(x)}

, where

g(x) = 2x + 1

and

g'(x) = 2

.

3

Sum the derivatives to find the complete expression for

\frac{dy}{dx}

.

\frac{dy}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x + \frac{2}{2x + 1}

The derivative of a sum is equal to the sum of the derivatives.

4

Evaluate the derivative at

x = 0

.

\frac{dy}{dx}\Big|_{x=0} = -e^0(0) + e^0(1)^2 + \frac{2}{1} = 3

Since

\tan(0) = 0

,

e^0 = 1

, and

\sec(0) = 1

, substituting

x = 0

simplifies the derivative to

0 + 1 + 2 = 3

.

Anahtar Kavram

Differentiation of Transcendental Functions (Product and Chain Rules)

Soru 139Soru →

What is the value of the definite integral $\int_{0}^{4} \sqrt{x} \, dx$ ?

\frac{16}{3}

8

12

\frac{8}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{16}{3}$

Cevap

\frac{16}{3}

Applying the power rule to

x^{1/2}

gives

\frac{2}{3}x^{3/2}

. Substituting the upper limit

x = 4

yields

\frac{2}{3}(4)^{3/2} = \frac{2}{3}(8) = \frac{16}{3}

, and substituting the lower limit

x = 0

yields 0. Subtracting the lower limit result from the upper limit result gives

\frac{16}{3}

.

Adım Adım Çözüm

1

Express the integrand with a fractional exponent

\sqrt{x} = x^{1/2}

Rewriting the square root as a fractional exponent allows the application of the power rule of integration.

2

Find the antiderivative using the power rule of integration

\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}

The power rule states that

\int x^n \, dx = \frac{x^{n+1}}{n+1}

for

n \neq -1

.

3

Evaluate the antiderivative at the limits 4 and 0

\left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(8) - 0 = \frac{16}{3}

By the Fundamental Theorem of Calculus, evaluate

F(b) - F(a)

.

Anahtar Kavram

Definite Integration using the Power Rule

Soru 140Soru →

Find the total area of the region bounded by the curve $y = 3x^2 - 6x$ , the $x$ -axis, and the vertical lines $x = 0$ and $x = 3$ .

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The total area bounded by the curve and the x-axis between x = 0 and x = 3 is 8 square units.

To find the total geometric area bounded by a curve and the x-axis, we must split the integral at any x-intercepts within the domain. For

y = 3x^2 - 6x

, the x-intercepts are

x = 0

and

x = 2

. Between

x = 0

and

x = 2

, the curve lies below the x-axis, giving an area magnitude of

4

. Between

x = 2

and

x = 3

, the curve lies above the x-axis, giving an area magnitude of

4

. Summing these positive magnitudes gives a total area of

8

.

Adım Adım Çözüm

1

Find the roots of the curve

y = 3x^2 - 6x

within the given interval

[0, 3]

.

Setting

3x^2 - 6x = 0

yields

3x(x - 2) = 0

, giving

x = 0

and

x = 2

.

Roots inside the integration boundaries indicate where the curve crosses the x-axis, changing the sign of

y

.

2

Determine the position of the curve relative to the x-axis on each sub-interval.

On

[0, 2]

,

y \le 0

(below the x-axis). On

[2, 3]

,

y \ge 0

(above the x-axis).

Geometric area must be non-negative, so regions below the x-axis require integrating

-y

or taking the absolute value of the integral.

3

Evaluate the area

A_1

for the region below the x-axis from

x = 0

to

x = 2

.

A_1 = \int_{0}^{2} (6x - 3x^2) \, dx = \left[ 3x^2 - x^3 \right]_{0}^{2} = (3(4) - 8) - 0 = 4

.

Integrating

-y = 6x - 3x^2

yields the positive magnitude of the area below the x-axis.

4

Evaluate the area

A_2

for the region above the x-axis from

x = 2

to

x = 3

.

A_2 = \int_{2}^{3} (3x^2 - 6x) \, dx = \left[ x^3 - 3x^2 \right]_{2}^{3} = (3^3 - 3(3^2)) - (2^3 - 3(2^2)) = 0 - (-4) = 4

.

Direct integration of

y

on

[2, 3]

gives the area above the x-axis.

5

Combine the areas of both sub-regions.

Total Area

= A_1 + A_2 = 4 + 4 = 8

.

The total geometric area is the sum of the magnitudes of the areas of all separate bounded regions.

Anahtar Kavram

Calculating area under curves crossing the x-axis by splitting definite integrals at real roots

Calculus

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