Calculus Practice Questions — JAMB UTME — Page 6

Evaluate the limit:

\lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2}

What is the numerical value of this limit?

Show answer & explanation

Answer: 48

Answer

The numerical value of the limit is 48.

Evaluating the limit of

\frac{x^3 - 8}{\sqrt{x + 2} - 2}

as

x \to 2

gives an indeterminate form

\frac{0}{0}

. Factorizing the numerator gives

(x - 2)(x^2 + 2x + 4)

, and rationalizing the denominator by multiplying numerator and denominator by

(\sqrt{x + 2} + 2)

converts the denominator to

x - 2

. Canceling

(x - 2)

leaves

(x^2 + 2x + 4)(\sqrt{x + 2} + 2)

. Evaluating at

x = 2

gives

(4 + 4 + 4)(\sqrt{4} + 2) = 12 \times 4 = 48

.

Step-by-Step Solution

1

Identify the limit form via direct substitution

Substituting

x = 2

yields

\frac{0}{0}

.

Direct evaluation results in an indeterminate form, requiring algebraic manipulation to eliminate the zero factor.

2

Factorize the numerator using the difference of cubes formula

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Exposing the factor

(x - 2)

is essential to resolving the zero denominator.

3

Rationalize the denominator using its algebraic conjugate

Multiply top and bottom by

(\sqrt{x + 2} + 2)

to get denominator

(x + 2) - 4 = x - 2

.

Applying

(a-b)(a+b) = a^2 - b^2

eliminates the square root from the denominator.

4

Cancel the common factor and compute the final value

\lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x + 2} + 2) = (12)(4) = 48.

With

(x - 2)

cancelled for

x \neq 2

, direct substitution now yields a defined real number.

Key Concept

Limits of Indeterminate Forms using Difference of Cubes and Surd Rationalization

Estimated Time:2m 30s

Question 102Question →

If $y = 3e^{2x} + \ln(\cos x)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 6

Answer

The value of the derivative of

y = 3e^{2x} + \ln(\cos x)

at

x = 0

is 6.

Differentiating each transcendental term individually using the chain rule yields

\frac{dy}{dx} = 6e^{2x} - \tan x

. Substituting

x = 0

gives

6e^0 - \tan 0 = 6 - 0 = 6

.

Step-by-Step Solution

1

Differentiate the exponential term

3e^{2x}

6e^{2x}

Applying the derivative rule for exponential functions

\frac{d}{dx}[a e^{kx}] = a k e^{kx}

.

2

Differentiate the logarithmic term

\ln(\cos x)

-\tan x

Applying the chain rule

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

where

u = \cos x

gives

\frac{-\sin x}{\cos x} = -\tan x

.

3

Combine terms and evaluate at

x = 0

6

Substituting

x = 0

into

\frac{dy}{dx} = 6e^{2x} - \tan x

yields

6e^0 - \tan 0 = 6(1) - 0 = 6

.

Key Concept

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

Question 103Question →

If $y = 2\sin(3x) + e^{4x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Show answer & explanation

Answer: 10

Answer

10

Differentiating

y = 2\sin(3x) + e^{4x}

yields

\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}

. Evaluating this expression at

x = 0

gives

6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10

.

Step-by-Step Solution

1

Differentiate each term of

y = 2\sin(3x) + e^{4x}

with respect to

x

.

\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}

Applying the chain rule gives

\frac{d}{dx}[2\sin(3x)] = 2 \times 3\cos(3x) = 6\cos(3x)

and

\frac{d}{dx}[e^{4x}] = 4e^{4x}

.

2

Evaluate the derivative at

x = 0

.

6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10

Substituting

x = 0

gives

\cos(0) = 1

and

e^{0} = 1

.

Key Concept

Differentiation of trigonometric and exponential functions

Question 104Question →

At what point on the curve $y = 2x^2 - 8x + 5$ is the normal line parallel to the straight line $x + 4y - 7 = 0$ ?

(3, -1)

(1, -1)

(3, 11)

(4, 5)

Show answer & explanation

Answer: $(3, -1)$

Answer

The point of contact on the curve is

(3, -1)

.

The line

x + 4y - 7 = 0

has a slope of

-\frac{1}{4}

. Since the normal line is parallel to this line, the normal gradient is

m_n = -\frac{1}{4}

. Consequently, the tangent gradient must be

m_t = 4

because

m_t \cdot m_n = -1

. Equating the derivative

\frac{dy}{dx} = 4x - 8

to

4

yields

x = 3

. Substituting

x = 3

into the curve equation

y = 2x^2 - 8x + 5

gives

y = -1

, yielding the point

(3, -1)

.

Step-by-Step Solution

1

Find the gradient of the given straight line.

Rearranging

x + 4y - 7 = 0

into slope-intercept form gives

y = -\frac{1}{4}x + \frac{7}{4}

, so the line's gradient is

m = -\frac{1}{4}

.

Parallel lines have equal gradients, so the normal line to the curve must have gradient

m_n = -\frac{1}{4}

.

2

Determine the required gradient of the tangent line.

Since

m_t \cdot m_n = -1

, we have

m_t = -\frac{1}{-\frac{1}{4}} = 4

.

The tangent and normal lines are perpendicular to each other.

3

Differentiate the curve's equation to find the

x

-coordinate.

\frac{dy}{dx} = 4x - 8

. Setting

\frac{dy}{dx} = 4

gives

4x - 8 = 4 \implies 4x = 12 \implies x = 3

.

The derivative represents the slope of the tangent line at any point

x

.

4

Substitute

x = 3

back into the curve's equation to find

y

.

y = 2(3)^2 - 8(3) + 5 = 2(9) - 24 + 5 = 18 - 24 + 5 = -1

.

The point of contact lies on the original curve.

Key Concept

Relationship between gradients of parallel lines, tangent lines, and normal lines to a curve

Question 105Question →

Evaluate the limit:

\lim_{x \to \infty} (\sqrt{x^2 + 6x} - x)

What is the numerical value of this limit?

3

6

0

Undefined

Show answer & explanation

Answer: 3

Answer

3

To evaluate the limit of

\sqrt{x^2 + 6x} - x

as

x \to \infty

, multiply and divide by its conjugate

\sqrt{x^2 + 6x} + x

. The numerator simplifies to

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

. Dividing both the numerator and denominator by

x

yields

\frac{6}{\sqrt{1 + 6/x} + 1}

. Taking the limit as

x \to \infty

reduces

\frac{6}{x}

to

0

, resulting in

\frac{6}{\sqrt{1} + 1} = 3

.

Step-by-Step Solution

1

Identify the indeterminate form

Direct evaluation gives

\infty - \infty

, which is an indeterminate form.

Substitution cannot be applied directly when subtracting infinite limits.

2

Multiply and divide by the algebraic conjugate

\lim_{x \to \infty} \frac{(\sqrt{x^2 + 6x} - x)(\sqrt{x^2 + 6x} + x)}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{6x}{\sqrt{x^2 + 6x} + x}

The identity

(a-b)(a+b) = a^2 - b^2

eliminates the square root in the numerator.

3

Factor

x

out of the denominator

\lim_{x \to \infty} \frac{6x}{x \left(\sqrt{1 + \frac{6}{x}} + 1\right)} = \lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1}

Dividing the numerator and denominator by

x

allows evaluation at infinity.

4

Compute the limit as

x \to \infty

Since

\lim_{x \to \infty} \frac{6}{x} = 0

, the expression becomes

\frac{6}{\sqrt{1 + 0} + 1} = \frac{6}{2} = 3

.

Terms with

x

in the denominator approach zero as

x

grows arbitrarily large.

Key Concept

Limits at infinity involving radical indeterminate forms of type

\infty - \infty

Estimated Time:2m 0s

Question 106Question →

Using differentiation from first principles, what is the derivative of the function $f(x) = x^2 - 5x$ with respect to $x$ ?

2x - 5

2x + 5

x - 5

2x

Show answer & explanation

Answer: $2x - 5$

Answer

The derivative of

f(x) = x^2 - 5x

with respect to

x

is

2x - 5

.

Evaluating the difference quotient

\frac{f(x+h)-f(x)}{h}

for

f(x) = x^2 - 5x

gives

\frac{x^2+2xh+h^2-5x-5h-(x^2-5x)}{h} = 2x + h - 5

. Taking the limit as

h \to 0

yields

2x - 5

.

Step-by-Step Solution

1

Write down the definition of the derivative from first principles.

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

This is the standard formula for differentiation from first principles.

2

Substitute

f(x) = x^2 - 5x

and

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

into the definition.

\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 5(x+h)] - (x^2 - 5x)}{h}

This sets up the difference quotient for the given polynomial function.

3

Expand the numerator and combine like terms.

\frac{x^2 + 2xh + h^2 - 5x - 5h - x^2 + 5x}{h} = \frac{2xh + h^2 - 5h}{h} = 2x + h - 5

Expanding allows

x^2

and

-5x

terms to cancel out, leaving terms containing

h

.

4

Evaluate the limit as

h \to 0

.

\lim_{h \to 0} (2x + h - 5) = 2x - 5

As

h

approaches 0, the term

h

vanishes, giving the final derivative.

Key Concept

Differentiation from First Principles

Question 107Question →

If $y = e^{2x} \ln(e^x + \sin x)$ , determine the value of $\frac{d^2 y}{dx^2}$ at $x = 0$ .

Show answer & explanation

Answer: 5

Answer

5

Evaluating the second derivative of

y = e^{2x} \ln(e^x + \sin x)

at

x = 0

yields

5

. This is calculated by applying the product, chain, and quotient rules to get

\frac{d^2 y}{dx^2}

, and substituting

x = 0

, where

e^0 = 1

,

\sin 0 = 0

,

\cos 0 = 1

, and

\ln 1 = 0

.

Step-by-Step Solution

1

Differentiate

y = e^{2x} \ln(e^x + \sin x)

with respect to

x

using the product rule.

\frac{dy}{dx} = 2e^{2x} \ln(e^x + \sin x) + e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

The function is composed of

u(x) = e^{2x}

and

v(x) = \ln(e^x + \sin x)

. By chain rule,

v'(x) = \frac{e^x + \cos x}{e^x + \sin x}

.

2

Differentiate the first term

T_1(x) = 2e^{2x} \ln(e^x + \sin x)

to get

T_1'(x)

.

T_1'(x) = 4e^{2x} \ln(e^x + \sin x) + 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

Applying the product rule to

2e^{2x}

and

\ln(e^x + \sin x)

.

3

Differentiate the second term

T_2(x) = e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

using the product and quotient rules.

T_2'(x) = 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right) + e^{2x} \left(\frac{(e^x - \sin x)(e^x + \sin x) - (e^x + \cos x)^2}{(e^x + \sin x)^2}\right)

The derivative of

\frac{e^x + \cos x}{e^x + \sin x}

requires the quotient rule.

4

Evaluate

T_1'(0)

and

T_2'(0)

at

x = 0

.

T_1'(0) = 4(1)(0) + 2(1)(2) = 4

, and

T_2'(0) = 2(1)(2) + 1 \left(\frac{(1)(1) - 2^2}{1^2}\right) = 4 - 3 = 1

.

At

x = 0

,

e^0 = 1

,

\sin 0 = 0

,

\cos 0 = 1

, and

\ln(1) = 0

.

5

Sum the evaluated derivative components to find

\frac{d^2 y}{dx^2}\Big|_{x=0}

.

\frac{d^2 y}{dx^2}\Big|_{x=0} = T_1'(0) + T_2'(0) = 4 + 1 = 5

Combining the evaluated terms gives the final numerical value.

Key Concept

Higher-Order Derivatives of Combined Transcendental Functions

Question 108Question →

What is the numerical value of the limit:

\lim_{x \to 0} \frac{1 - \cos(4x)}{x \sin(2x)}

?

4

2

8

0

Show answer & explanation

Answer: 4

Answer

The numerical value of the limit is 4.

Applying the double-angle trigonometric identity

1 - \cos(4x) = 2\sin^2(2x)

reduces the expression to

\frac{2\sin(2x)}{x}

. Rewriting this as

4 \cdot \frac{\sin(2x)}{2x}

and taking the limit as

x \to 0

using the standard limit

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

yields 4.

Step-by-Step Solution

1

Identify the form of the limit

Direct substitution of

x = 0

gives

\frac{1 - \cos(0)}{0 \cdot \sin(0)} = \frac{0}{0}

, which is an indeterminate form.

Indeterminate forms require algebraic simplification or trigonometric identities before evaluating the limit.

2

Apply trigonometric identity

Use

1 - \cos(4x) = 2\sin^2(2x)

to rewrite the numerator.

This transforms the numerator into a form containing sine terms matching the denominator.

3

Simplify the algebraic expression

\lim_{x \to 0} \frac{2\sin^2(2x)}{x\sin(2x)} = \lim_{x \to 0} \frac{2\sin(2x)}{x}

Cancel out the common

\sin(2x)

factor from numerator and denominator for

x \neq 0

.

4

Evaluate using the standard trigonometric limit

\lim_{x \to 0} 4 \cdot \frac{\sin(2x)}{2x} = 4 \cdot 1 = 4

Since

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

, setting

\theta = 2x

gives

\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1

.

Key Concept

Limits of Trigonometric Functions and Indeterminate Forms

Question 109Question →

What is the derivative of the function $f(x) = 3x^2 + 5x$ with respect to $x$ , obtained using differentiation from first principles?

6x + 5

6x - 5

3x + 5

6x + 5 + 3h

Show answer & explanation

Answer: $6x + 5$

Answer

The derivative of the function is

6x + 5

.

Using the first-principles formula

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

, we expand

f(x+h) = 3(x+h)^2 + 5(x+h) = 3x^2 + 6xh + 3h^2 + 5x + 5h

. Subtracting

f(x) = 3x^2 + 5x

yields

6xh + 3h^2 + 5h

. Dividing by

h

produces

6x + 3h + 5

. Taking the limit as

h \to 0

gives

6x + 5

.

Step-by-Step Solution

1

Evaluate

f(x+h)

for

f(x) = 3x^2 + 5x

f(x+h) = 3(x+h)^2 + 5(x+h) = 3(x^2 + 2xh + h^2) + 5x + 5h = 3x^2 + 6xh + 3h^2 + 5x + 5h

Substitute

(x+h)

into the original function expression and expand algebraically.

2

Find the difference

f(x+h) - f(x)

f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 5x + 5h) - (3x^2 + 5x) = 6xh + 3h^2 + 5h

Subtract the original function

f(x)

to find the net change in output.

3

Divide the difference by

h

to set up the difference quotient

\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 + 5h}{h} = 6x + 3h + 5

Cancel out

h

from each term in the numerator.

4

Take the limit as

h \to 0

f'(x) = \lim_{h \to 0} (6x + 3h + 5) = 6x + 5

Evaluate the expression as

h

approaches 0 to determine the instantaneous rate of change.

Key Concept

Differentiation from First Principles

Estimated Time:1m 0s

Question 110Question →

What is the $y$ -intercept of the tangent line to the curve $y = x^3 - 2x + 4$ at the point where $x = 1$ ?

Show answer & explanation

Answer: 2

Answer

The y-intercept of the tangent line is 2.

The curve evaluated at

x=1

gives point

(1,3)

. The derivative

y'=3x^2-2

gives slope

m=1

at

x=1

. The tangent line equation is

y-3=1(x-1)

, which simplifies to

y=x+2

. The

y

-intercept occurs at

x=0

, giving

y=2

.

Step-by-Step Solution

1

Find the y-coordinate of the point on the curve at

x = 1

.

At

x = 1

,

y = (1)^3 - 2(1) + 4 = 3

. The point of tangency is

(1, 3)

.

The point of tangency lies on the curve.

2

Find the gradient function of the curve by differentiation.

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

.

The first derivative represents the slope of the tangent line.

3

Calculate the slope of the tangent line at

x = 1

.

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

.

Substituting

x = 1

into the derivative yields the slope at that specific point.

4

Formulate the equation of the tangent line.

y - 3 = 1(x - 1) \implies y = x + 2

.

Use point-slope form

y - y_1 = m(x - x_1)

with point

(1,3)

and slope

m=1

.

5

Find the

y

-intercept of the tangent line.

Setting

x = 0

gives

y = 2

.

The

y

-intercept is the value of

y

where the line crosses the vertical axis.

Key Concept

Tangents and Normals to Curves

Question 111Question →

What is the indefinite integral $\int (6x^5 + 8\sin(4x)) \, dx$ ?

x^6 - 2\cos(4x) + C

x^6 - 2\cos(4x)

x^6 + 2\cos(4x) + C

x^6 - 8\cos(4x) + C

Show answer & explanation

Answer: $x^6 - 2\cos(4x) + C$

Answer

x^6 - 2\cos(4x) + C

Integrating

6x^5

using the power rule gives

\frac{6x^6}{6} = x^6

. Integrating

8\sin(4x)

gives

8 \left(-\frac{\cos(4x)}{4}\right) = -2\cos(4x)

. Summing these and appending the constant of integration

C

yields

x^6 - 2\cos(4x) + C

.

Step-by-Step Solution

1

Split the integral into two individual terms

\int (6x^5 + 8\sin(4x)) \, dx = \int 6x^5 \, dx + \int 8\sin(4x) \, dx

Linearity property of integration allows term-by-term integration.

2

Integrate the polynomial term

6x^5

\int 6x^5 \, dx = 6 \cdot \frac{x^{5+1}}{5+1} = \frac{6x^6}{6} = x^6

Apply the power rule of integration:

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

.

3

Integrate the trigonometric term

8\sin(4x)

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

Apply the standard integral rule:

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

.

4

Combine results and add the constant of integration

x^6 - 2\cos(4x) + C

Indefinite integration requires an arbitrary constant

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:1m 30s

Question 112Question →

Using differentiation from first principles, what is the derivative $\frac{dy}{dx}$ of the function $y = 3 - 2x^2$ ?

-4x

4x

-2x

-4x - 2

Show answer & explanation

Answer: $-4x$

Answer

-4x

Applying the definition of derivative

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

to

f(x) = 3 - 2x^2

gives

\frac{-4xh - 2h^2}{h} = -4x - 2h

. Taking the limit as

h \to 0

results in

-4x

.

Step-by-Step Solution

1

Express

f(x+h)

for the function

f(x) = 3 - 2x^2

f(x+h) = 3 - 2(x+h)^2 = 3 - 2(x^2 + 2xh + h^2) = 3 - 2x^2 - 4xh - 2h^2

Substitute

(x+h)

into the place of

x

and expand the squared binomial.

2

Set up the difference quotient

\frac{f(x+h) - f(x)}{h}

\frac{(3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2)}{h} = \frac{-4xh - 2h^2}{h}

Subtract

f(x)

from

f(x+h)

to isolate terms involving increment

h

.

3

Divide numerator terms by

h

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

Factor out

h

to cancel the denominator.

4

Take the limit as

h \to 0

\lim_{h \to 0} (-4x - 2h) = -4x

Evaluating the limit gives the exact derivative function

\frac{dy}{dx}

.

Key Concept

Differentiation from First Principles

Question 113Question →

If $y = 5e^{2x} + \cos(4x)$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 10

Answer

10

Differentiating

y = 5e^{2x} + \cos(4x)

with respect to

x

yields

\frac{dy}{dx} = 10e^{2x} - 4\sin(4x)

. Substituting

x = 0

gives

10e^0 - 4\sin(0) = 10(1) - 0 = 10

.

Step-by-Step Solution

1

Differentiate

y = 5e^{2x} + \cos(4x)

with respect to

x

.

\frac{dy}{dx} = 10e^{2x} - 4\sin(4x)

The derivative of

e^{ax}

is

a e^{ax}

and the derivative of

\cos(ax)

is

-a \sin(ax)

.

2

Evaluate the derivative at

x = 0

.

10

Substitute

x = 0

into

10e^{2x} - 4\sin(4x)

to obtain

10(1) - 4(0) = 10

.

Key Concept

Differentiation of exponential and trigonometric functions

Question 114Question →

What is the $x$ -coordinate of the stationary point of the curve $y = 3x^2 - 12x + 7$ ?

2

-2

4

7

Show answer & explanation

Answer: $2$

Answer

The

x

-coordinate of the stationary point is

2

.

To find the stationary point, take the derivative of

y = 3x^2 - 12x + 7

with respect to

x

, obtaining

\frac{dy}{dx} = 6x - 12

. Setting the derivative equal to zero gives

6x - 12 = 0

, which solves to

x = 2

.

Step-by-Step Solution

1

Differentiate the equation of the curve with respect to

x

\frac{dy}{dx} = 6x - 12

Stationary points occur where the first derivative (gradient) of the function is equal to zero.

2

Set the derivative to zero and solve for

x

6x - 12 = 0 \implies 6x = 12 \implies x = 2

Solving this linear equation gives the exact value of

x

at which the tangent to the curve is horizontal.

Key Concept

Stationary Points of a Curve

Estimated Time:45s

Question 115Question →

If $y = \frac{e^{2x}}{\cos x}$ , calculate the value of the derivative $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 2

Answer

The value of the derivative

\frac{dy}{dx}

at

x = 0

is 2.

Applying the quotient rule to

y = \frac{e^{2x}}{\cos x}

yields

\frac{dy}{dx} = \frac{2e^{2x}\cos x + e^{2x}\sin x}{\cos^2 x}

. Substituting

x = 0

gives

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

.

Step-by-Step Solution

1

Set up the quotient rule components for

y = \frac{u(x)}{v(x)}

.

Let

u(x) = e^{2x}

and

v(x) = \cos x

.

The given function is a ratio of exponential and trigonometric functions.

2

Compute individual derivatives

\frac{du}{dx}

and

\frac{dv}{dx}

.

\frac{du}{dx} = 2e^{2x}

and

\frac{dv}{dx} = -\sin x

.

Using the chain rule for exponential functions and standard derivative rules for trigonometric functions.

3

Apply the quotient rule formula

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

.

\frac{dy}{dx} = \frac{\cos x (2e^{2x}) - e^{2x}(-\sin x)}{\cos^2 x} = \frac{e^{2x}(2\cos x + \sin x)}{\cos^2 x}

.

Combining terms gives the exact derivative function.

4

Evaluate the derivative at

x = 0

.

\frac{dy}{dx}\Big|_{x=0} = \frac{e^{0}(2\cos 0 + \sin 0)}{\cos^2 0} = \frac{1 \cdot (2 + 0)}{1} = 2

.

Using trigonometric and exponential values at zero:

e^0 = 1

,

\cos 0 = 1

, and

\sin 0 = 0

.

Key Concept

Differentiation of Exponential and Trigonometric Functions using Quotient Rule

Question 116Question →

A closed cylindrical metal container has a total surface area of $54\pi\text{ cm}^2$ . What radius, in centimeters, of the circular base will yield the maximum volume for the container?

Show answer & explanation

Answer: 3

Answer

The radius of the circular base that maximizes the volume is 3 cm.

To find the radius that yields maximum volume, we first express height

h

in terms of radius

r

using the total surface area formula

2\pi r^2 + 2\pi rh = 54\pi

, giving

h = \frac{27 - r^2}{r}

. Substituting this into the volume equation

V = \pi r^2 h

gives

V(r) = 27\pi r - \pi r^3

. Setting the first derivative

\frac{dV}{dr} = 27\pi - 3\pi r^2

to zero yields

3\pi r^2 = 27\pi

, so

r^2 = 9

and

r = 3\text{ cm}

. The second derivative

\frac{d^2V}{dr^2} = -6\pi r

evaluated at

r = 3

is

-18\pi

, which is strictly negative, confirming that

r = 3\text{ cm}

maximizes volume.

Step-by-Step Solution

1

Set up the surface area equation for a closed cylinder with the given value.

2\pi r^2 + 2\pi r h = 54\pi

A closed cylinder consists of two circular bases (

2\pi r^2

) and a curved lateral surface (

2\pi r h

).

2

Express

h

in terms of

r

.

h = \frac{27 - r^2}{r}

Dividing the surface area equation by

2\pi

yields

r^2 + rh = 27

, allowing

h

to be isolated.

3

Substitute

h

into the volume formula

V = \pi r^2 h

to write volume as a function of

r

only.

V(r) = 27\pi r - \pi r^3

To maximize volume using calculus, the volume equation must be expressed in terms of a single variable.

4

Differentiate

V(r)

with respect to

r

and set the derivative equal to zero to find stationary points.

\frac{dV}{dr} = 27\pi - 3\pi r^2 = 0 \implies r = 3

Maximum volume occurs at a stationary point where the first derivative is zero.

5

Verify that

r = 3

produces a maximum using the second derivative test.

\frac{d^2V}{dr^2} = -6\pi r

; at

r = 3

,

\frac{d^2V}{dr^2} = -18\pi < 0

A negative second derivative indicates a local maximum.

Key Concept

Optimization and Stationary Points in Mensuration

Estimated Time:2m 0s

Question 117Question →

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

3\sin(6x)e^{\sin^2(3x)}

-3\sin(6x)e^{\sin^2(3x)}

\sin(6x)e^{\sin^2(3x)}

6\cos(3x)e^{\sin^2(3x)}

Show answer & explanation

Answer: $3\sin(6x)e^{\sin^2(3x)}$

Answer

The derivative

\frac{dy}{dx}

is

3\sin(6x)e^{\sin^2(3x)}

.

Differentiating

y = e^{\sin^2(3x)}

requires applying the chain rule step-by-step: first differentiating the exponential function to get

e^{\sin^2(3x)}

, then differentiating

\sin^2(3x)

to obtain

2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x)

. Multiplying these together and applying the double-angle identity

2\sin(3x)\cos(3x) = \sin(6x)

yields

3\sin(6x)e^{\sin^2(3x)}

.

Step-by-Step Solution

1

Identify the inner function

u

and outer function

y

for applying the chain rule.

Let

u = \sin^2(3x) = (\sin(3x))^2

, so

y = e^u

.

The function is an exponential function whose exponent is a composite trigonometric function.

2

Differentiate

u = (\sin(3x))^2

with respect to

x

using the chain rule.

\frac{du}{dx} = 2\sin(3x) \cdot \frac{d}{dx}(\sin(3x)) = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x).

The derivative of

[g(x)]^2

is

2g(x)g'(x)

, and

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

.

3

Differentiate

y = e^u

with respect to

x

using

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

\frac{dy}{dx} = e^{\sin^2(3x)} \cdot 6\sin(3x)\cos(3x).

The derivative of

e^u

with respect to

u

is

e^u

.

4

Simplify the expression using the trigonometric double-angle identity

2\sin\theta\cos\theta = \sin(2\theta)

, where

\theta = 3x

.

\frac{dy}{dx} = 3 \cdot (2\sin(3x)\cos(3x)) e^{\sin^2(3x)} = 3\sin(6x)e^{\sin^2(3x)}.

Rewriting

6\sin(3x)\cos(3x)

as

3\sin(6x)

simplifies the expression to standard examination form.

Key Concept

Chain Rule for Composite Exponential and Trigonometric Functions

Estimated Time:2m 0s

Question 118Question →

If $y = \int (12x^2 + 6\sin(2x)) \, dx$ and $y = 10$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Show answer & explanation

Answer: 13

Answer

The value of the constant of integration

C

is

13

.

Integrating

12x^2 + 6\sin(2x)

gives

y = 4x^3 - 3\cos(2x) + C

. Substituting

x = 0

yields

y(0) = 4(0)^3 - 3\cos(0) + C = -3 + C

. Setting

-3 + C = 10

and solving for

C

gives

C = 13

.

Step-by-Step Solution

1

Integrate the function with respect to

x

y = 4x^3 - 3\cos(2x) + C

Applying the power rule

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

and trigonometric integration rule

\int k\sin(bx) \, dx = -\frac{k}{b}\cos(bx)

.

2

Apply the initial condition

x = 0

and

y = 10

10 = 4(0)^3 - 3\cos(0) + C

Substituting the boundary values to solve for the specific constant of integration.

3

Evaluate trigonometric function and solve for

C

C = 13

Since

\cos(0) = 1

, the equation simplifies to

10 = -3 + C

, leading directly to

C = 13

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

Question 119Question →

Using differentiation from first principles, what is the derivative of the function $f(x) = 4 - x^2$ with respect to $x$ ?

-2x

2x

-2x - 1

4 - 2x

Show answer & explanation

Answer: $-2x$

Answer

-2x

Differentiating from first principles involves finding the limit of

\frac{f(x+h) - f(x)}{h}

as

h \to 0

. For

f(x) = 4 - x^2

, expanding

f(x+h)

gives

4 - x^2 - 2xh - h^2

. Subtracting

f(x)

yields

-2xh - h^2

, and dividing by

h

gives

-2x - h

. Taking the limit as

h \to 0

leaves

-2x

.

Step-by-Step Solution

1

Express

f(x+h)

for

f(x) = 4 - x^2

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

Substitute

x+h

into the original function definition.

2

Set up the difference

f(x+h) - f(x)

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

Subtract

f(x)

to find the net change in

y

.

3

Divide the difference by

h

to form the difference quotient

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

Divide each term in the numerator by

h

.

4

Take the limit as

h \to 0

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

Evaluate the derivative by letting

h

approach zero.

Key Concept

Differentiation from First Principles

Question 120Question →

What is the gradient of the normal line to the curve $y = x^2 - 3x + 5$ at the point where $x = 1$ ?

Show answer & explanation

Answer: 1

Answer

The gradient of the normal line to the curve at

x = 1

is 1.

Differentiating

y = x^2 - 3x + 5

gives

\frac{dy}{dx} = 2x - 3

. Evaluating this derivative at

x = 1

gives the tangent gradient

m_t = -1

. Because the normal line is perpendicular to the tangent, its gradient is

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

.

Step-by-Step Solution

1

Differentiate the function with respect to x

\frac{dy}{dx} = 2x - 3

The derivative of a function gives the slope of the tangent line at any given x-coordinate.

2

Evaluate the derivative at x = 1

m_t = -1

Substituting the given point's x-coordinate into the gradient function yields the slope of the tangent.

3

Calculate the negative reciprocal of the tangent slope

m_n = 1

Since the normal line is perpendicular to the tangent line, its gradient is m_n = -1 / m_t.

Key Concept

Gradient of a Normal Line

Calculus

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