Calculus Practice Questions — JAMB UTME

Question 1Question →

A curve is defined by the equation $y = x^2 - 5x + 6$ . What is the $x$ -intercept of the line normal to the curve at the point where $x = 1$ ?

Show answer & explanation

Answer: -5

Answer

The x-intercept of the normal line to the curve at x = 1 is -5.

Substituting

x = 1

into

y = x^2 - 5x + 6

yields

y = 2

, identifying the point

(1, 2)

. Differentiating gives

\frac{dy}{dx} = 2x - 5

, which equals

-3

at

x = 1

. The normal line gradient is the negative reciprocal,

\frac{1}{3}

. The line equation

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

simplifies to

x - 3y + 5 = 0

. Setting

y = 0

gives

x = -5

.

Step-by-Step Solution

1

Calculate the y-coordinate at x = 1 to determine the point of tangency

Substituting

x = 1

into

y = x^2 - 5x + 6

gives

y = (1)^2 - 5(1) + 6 = 2

, yielding the point

(1, 2)

.

The normal line intersects the curve at the point of tangency.

2

Find the derivative of the curve and evaluate the tangent slope

\frac{dy}{dx} = 2x - 5

. At

x = 1

,

m_t = 2(1) - 5 = -3

.

The derivative evaluated at a specific point gives the slope of the tangent line to the curve.

3

Compute the slope of the normal line

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

.

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

4

Construct the normal line equation and determine its x-intercept

Using point-slope form:

y - 2 = \frac{1}{3}(x - 1) \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0

. Setting

y = 0

gives

x + 5 = 0 \implies x = -5

.

The

x

-intercept occurs where the line crosses the

x

-axis, meaning

y = 0

.

Key Concept

The slope of the normal line to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative at that point,

m_n = -\frac{1}{f'(a)}

.

Question 2Question →

If $y = \ln(\cos 2x)$ , what is $\frac{dy}{dx}$ ?

-2\tan 2x

2\tan 2x

-\tan 2x

-2\cot 2x

Show answer & explanation

Answer: $-2\tan 2x$

Answer

-2\tan 2x

Applying the chain rule to

y = \ln(\cos 2x)

yields

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2\frac{\sin 2x}{\cos 2x} = -2\tan 2x

.

Step-by-Step Solution

1

Identify the inner and outer functions for the chain rule.

Let

u = \cos 2x

, so

y = \ln u

.

The given function

y = \ln(\cos 2x)

is a composite transcendental function.

2

Differentiate

y

with respect to

u

, and

u

with respect to

x

.

\frac{dy}{du} = \frac{1}{u} = \frac{1}{\cos 2x}

, and

\frac{du}{dx} = -2\sin 2x

.

The derivative of

\ln u

is

\frac{1}{u}

and the derivative of

\cos 2x

is

-2\sin 2x

using the chain rule.

3

Apply the chain rule

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

and simplify.

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-2\sin 2x) = -2\left(\frac{\sin 2x}{\cos 2x}\right) = -2\tan 2x

.

Using the trigonometric identity

\frac{\sin 2x}{\cos 2x} = \tan 2x

simplifies the expression into standard form.

Key Concept

Differentiation of composite logarithmic and trigonometric functions using the Chain Rule

Estimated Time:1m 0s

Question 3Question →

If $y = 2e^{4x} + \ln(2x + 1)$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Show answer & explanation

Answer: 10

Answer

10

Differentiating

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

yields

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

. Evaluating this derivative at

x = 0

gives

8(e^0) + \frac{2}{2(0) + 1} = 8(1) + 2 = 10

.

Step-by-Step Solution

1

Differentiate each term of

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

By the chain rule,

\frac{d}{dx}(ae^{kx}) = ak e^{kx}

and

\frac{d}{dx}(\ln(u(x))) = \frac{u'(x)}{u(x)}

.

2

Evaluate the derivative at

x = 0

\left.\frac{dy}{dx}\right|_{x=0} = 8e^{0} + \frac{2}{2(0) + 1} = 8(1) + \frac{2}{1} = 10

Substitute

x = 0

into the derived expression and simplify using

e^0 = 1

.

Key Concept

Differentiation of exponential and logarithmic functions using the chain rule

Question 4Question →

If $y = e^{2x} \tan x$ , what is $\frac{dy}{dx}$ ?

e^{2x}(2\tan x + \sec^2 x)

e^{2x}(\tan x + \sec^2 x)

e^{2x}(2\tan x - \sec^2 x)

2e^{2x} \sec^2 x

Show answer & explanation

Answer: $e^{2x}(2\tan x + \sec^2 x)$

Answer

e^{2x}(2\tan x + \sec^2 x)

Applying the product rule

\frac{d}{dx}[uv] = u'v + uv'

with

u = e^{2x}

and

v = \tan x

gives

u' = 2e^{2x}

and

v' = \sec^2 x

. Substituting these into the formula yields

2e^{2x}\tan x + e^{2x}\sec^2 x = e^{2x}(2\tan x + \sec^2 x)

.

Step-by-Step Solution

1

Identify the component functions for the product rule

Let

u(x) = e^{2x}

and

v(x) = \tan x

.

The given function

y = e^{2x} \tan x

is a product of two functions.

2

Differentiate each component function separately

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

by the chain rule, and

\frac{dv}{dx} = \sec^2 x

.

The derivative of

e^{kx}

is

k e^{kx}

and the derivative of

\tan x

is

\sec^2 x

.

3

Apply the product rule formula

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

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

.

Combining the terms according to the standard product rule.

4

Factor out the common term

e^{2x}

\frac{dy}{dx} = e^{2x}(2\tan x + \sec^2 x)

.

Simplifying to match standard exam response format.

Key Concept

Differentiation of Exponential and Trigonometric Functions using the Product Rule

Estimated Time:1m 30s

Question 5Question →

A curve is given by the equation $y = \frac{x}{x^2 + 4}$ . What is the maximum value of $y$ on this curve?

\frac{1}{4}

-\frac{1}{4}

2

0

Show answer & explanation

Answer: $\frac{1}{4}$

Answer

The maximum value of

y

on the curve is

\frac{1}{4}

.

To find the maximum value of

y = \frac{x}{x^2 + 4}

, differentiate

y

using the quotient rule to obtain

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting the numerator to zero gives

x = \pm 2

. Since

\frac{dy}{dx}

is positive for

x < 2

and negative for

x > 2

,

x = 2

corresponds to a local maximum. Substituting

x = 2

into

y = \frac{x}{x^2 + 4}

yields

y = \frac{2}{8} = \frac{1}{4}

.

Step-by-Step Solution

1

Differentiate

y = \frac{x}{x^2 + 4}

using the quotient rule

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative

\frac{dy}{dx}

equals zero.

2

Find the stationary points by setting

\frac{dy}{dx} = 0

4 - x^2 = 0 \implies x = 2 \text{ or } x = -2

A fraction equals zero when its numerator is zero.

3

Determine the nature of the stationary point at

x = 2

\text{For } x < 2, \frac{dy}{dx} > 0; \text{ for } x > 2, \frac{dy}{dx} < 0 \implies x = 2 \text{ is a maximum point}

The derivative changes sign from positive to negative across a maximum point.

4

Evaluate

y

at

x = 2

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

Substituting the

x

-coordinate into the curve equation yields the maximum value of

y

.

Key Concept

Stationary Points, Maxima, and Minima

Question 6Question →

A curve is defined by the equation $y = x^3 - 4x + 3$ . What is the $y$ -intercept of the tangent line to the curve at the point where $x = 2$ ?

Show answer & explanation

Answer: -13

Answer

The y-intercept of the tangent line is

-13

.

At

x = 2

, substituting into

y = x^3 - 4x + 3

yields

y = 3

. The derivative

\frac{dy}{dx} = 3x^2 - 4

evaluated at

x = 2

gives

m = 8

. The tangent line equation is

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

. Setting

x = 0

gives the

y

-intercept of

-13

.

Step-by-Step Solution

1

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

x = 2

y = 3

, giving the point

(2, 3)

The point of contact must lie on the curve.

2

Differentiate the curve equation to find the gradient function

\frac{dy}{dx} = 3x^2 - 4

The derivative of a function gives the gradient of the tangent at any point

x

.

3

Evaluate the derivative at

x = 2

to find the gradient of the tangent

m

m = 8

Substituting

x = 2

into

3x^2 - 4

yields

3(4) - 4 = 8

.

4

Formulate the linear equation of the tangent line using point

(2, 3)

and gradient

m = 8

y = 8x - 13

Applying

y - y_1 = m(x - x_1)

gives

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

.

5

Extract the y-intercept of the tangent line

y-intercept =

-13

In slope-intercept form

y = mx + c

, the constant term

c = -13

is the y-intercept.

Key Concept

Tangents to Curves and Line Intercepts

Question 7Question →

Find the gradient of the normal to the curve $y = \frac{x + 1}{x - 1}$ at the point where $x = 3$ .

Show answer & explanation

Answer: 2

Answer

The gradient of the normal to the curve at

x = 3

is

2

.

Differentiating

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

via the quotient rule gives

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

. At

x = 3

, the tangent gradient is

m_t = \frac{-2}{4} = -\frac{1}{2}

. Since the normal is perpendicular to the tangent, its gradient is

m_n = -\frac{1}{m_t} = 2

.

Step-by-Step Solution

1

Differentiate the rational function with respect to

x

Applying the quotient rule

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}

where

u = x + 1

and

v = x - 1

yields

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

.

The first derivative determines the slope function of the tangent line to the curve.

2

Substitute

x = 3

into the derivative to find the tangent slope

m_t

m_t = \frac{-2}{(3 - 1)^2} = \frac{-2}{4} = -\frac{1}{2}

.

Evaluating the derivative at the given

x

-coordinate provides the exact gradient of the tangent at that point.

3

Calculate the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-1/2} = 2

.

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

Key Concept

The slope of the normal line to a curve at a given point is the negative reciprocal of the slope of the tangent line at that point (

m_n = -\frac{1}{m_t}

).

Question 8Question →

A curve is defined by the equation $y = x^3 - 3x^2 + k$ , where $k$ is a constant. If the local minimum value of $y$ on the curve is $2$ , what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the constant

k

is

6

.

To find the constant

k

, differentiate the curve equation to obtain

\frac{dy}{dx} = 3x^2 - 6x

. Setting

\frac{dy}{dx} = 0

gives stationary points at

x = 0

and

x = 2

. Calculating the second derivative

\frac{d^2y}{dx^2} = 6x - 6

shows

\frac{d^2y}{dx^2} = 6 > 0

at

x = 2

, confirming that the local minimum occurs at

x = 2

. Substituting

x = 2

and the minimum value

y = 2

into

y = x^3 - 3x^2 + k

yields

2 = 8 - 12 + k

, which simplifies to

k = 6

.

Step-by-Step Solution

1

Differentiate

y = x^3 - 3x^2 + k

with respect to

x

.

\frac{dy}{dx} = 3x^2 - 6x

Stationary points occur where the first derivative equals zero.

2

Solve

\frac{dy}{dx} = 0

for

x

.

x = 0

or

x = 2

These

x

-values locate the turning points on the curve.

3

Evaluate the second derivative

\frac{d^2y}{dx^2} = 6x - 6

at each stationary point.

At

x = 0

,

\frac{d^2y}{dx^2} = -6

(local maximum); at

x = 2

,

\frac{d^2y}{dx^2} = 6

(local minimum).

A positive second derivative indicates a local minimum point.

4

Substitute

x = 2

and

y = 2

into the curve equation

y = x^3 - 3x^2 + k

.

2 = (2)^3 - 3(2)^2 + k \implies 2 = -4 + k \implies k = 6

The local minimum value of

y

is attained at

x = 2

.

Key Concept

Stationary Points, Maxima, and Minima

Question 9Question →

A curve is defined by the equation $y = 2x^3 - 9x^2 + 12x + 5$ . What is the value of $y$ at its maximum stationary point?

Show answer & explanation

Answer: 10

Answer

The value of

y

at the maximum stationary point is

10

.

To locate the maximum stationary point, set the first derivative

\frac{dy}{dx} = 6x^2 - 18x + 12

to zero, obtaining stationary values

x = 1

and

x = 2

. Testing in the second derivative

\frac{d^2y}{dx^2} = 12x - 18

gives

-6

at

x = 1

, confirming a local maximum. Substituting

x = 1

into the original cubic equation gives

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

.

Step-by-Step Solution

1

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 6x^2 - 18x + 12

Stationary points occur where the gradient of the curve is zero.

2

Set the first derivative to zero and solve for

x

.

x = 1 \text{ or } x = 2

Factoring

6(x - 1)(x - 2) = 0

yields the

x

-coordinates of the turning points.

3

Determine the nature of the stationary points using the second derivative test.

\frac{d^2y}{dx^2} = 12x - 18; \quad \text{at } x = 1, \frac{d^2y}{dx^2} = -6 < 0

A negative second derivative indicates a local maximum stationary point.

4

Substitute

x = 1

into the original function to determine

y

.

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

Evaluating the curve function at the maximum

x

-coordinate provides the corresponding maximum

y

-value.

Key Concept

Stationary Points, Maxima, and Minima

Question 10Question →

If $\int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C$ , where $a$ , $b$ , and $C$ are constants, what is the value of $a + b$ ?

Show answer & explanation

Answer: 4

Answer

The value of

a + b

is

4

.

Integrating

6x^2

gives

2x^3

, and integrating

-4\sin(2x)

gives

2\cos(2x)

. Comparing

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

to

ax^3 + b\cos(2x) + C

yields

a=2

and

b=2

, giving

a+b=4

.

Step-by-Step Solution

1

Integrate each term of the integrand

(6x^2 - 4\sin(2x))

with respect to

x

.

\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3

, and

\int -4\sin(2x) \, dx = -4 \cdot \left(-\frac{1}{2}\cos(2x)\right) = 2\cos(2x)

.

Applying the power rule for integration

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

and the standard trigonometric integral formula

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

.

2

Combine the calculated antiderivative terms and include the constant of integration

C

.

\int (6x^2 - 4\sin(2x)) \, dx = 2x^3 + 2\cos(2x) + C

.

Summing the individual term-by-term antiderivatives produces the complete indefinite integral.

3

Compare the resulting expression with

ax^3 + b\cos(2x) + C

to determine

a

and

b

.

a = 2

and

b = 2

.

Matching corresponding coefficients of

x^3

and

\cos(2x)

.

4

Compute

a + b

.

2 + 2 = 4

.

Evaluating the sum of the extracted coefficients.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:1m 30s

Question 11Question →

What is the indefinite integral $\int (4x^3 - 6\cos(3x)) \, dx$ ?

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

x^4 - 2\sin(3x)

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

x^4 - 18\sin(3x) + C

Show answer & explanation

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

Answer

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

Integrating term-by-term,

\int 4x^3 \, dx = x^4

and

\int -6\cos(3x) \, dx = -2\sin(3x)

. Summing these and including the arbitrary constant

C

produces

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

.

Step-by-Step Solution

1

Integrate the polynomial term

4x^3

using the power rule

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

.

\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4

The power rule for integration adds

1

to the exponent and divides by the new exponent.

2

Integrate the trigonometric term

-6\cos(3x)

using

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

.

\int -6\cos(3x) \, dx = -6 \cdot \frac{1}{3}\sin(3x) = -2\sin(3x)

Integrating cosine gives positive sine, divided by the coefficient of

x

.

3

Combine the results and append the constant of integration

C

.

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

An indefinite integral represents a family of functions and requires the constant of integration

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 12Question →

Using the first principles of differentiation for the reciprocal function $f(x) = \frac{4}{x}$ , evaluate the limit of the difference quotient as $h \to 0$ : $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . What is the simplified expression for the derivative $\frac{dy}{dx}$ ?

Show answer & explanation

Answer: -\frac{4}{x^2}; -4/x^2; -4 / x^2; -\frac{4}{x^{2}}

Answer

The derivative

\frac{dy}{dx}

is

-\frac{4}{x^2}

.

Substituting

f(x) = \frac{4}{x}

into the first principles limit formula yields

\lim_{h \to 0} \frac{4x - 4(x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-4h}{h \cdot x(x+h)}

. Canceling

h

gives

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

.

Step-by-Step Solution

1

Write the first principles formula and substitute

f(x) = \frac{4}{x}

.

\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}

Apply the definition of differentiation from first principles.

2

Combine the fractions in the numerator using a common denominator

x(x+h)

.

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

Simplify the numerator into a single fractional expression.

3

Divide the simplified numerator by

h

and cancel the common factor of

h

.

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

Eliminate the indeterminate factor of

h

from the denominator.

4

Evaluate the limit as

h \to 0

.

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

Substitute

h = 0

into the simplified algebraic expression.

Key Concept

Differentiation from First Principles

Question 13Question →

In differentiating the quadratic function $f(x) = 2x^2 + 5x$ from first principles, which of the following expressions represents the fully simplified difference quotient $\frac{f(x+h) - f(x)}{h}$ prior to evaluating the limit as $h \to 0$ ?

4x + 2h + 5

4x + h + 5

4x - 5 + 2h

2x + 2h + 5

Show answer & explanation

Answer: $4x + 2h + 5$

Answer

The fully simplified difference quotient is

4x + 2h + 5

.

Expanding

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

gives

2x^2 + 4xh + 2h^2 + 5x + 5h

. Subtracting

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

leaves

4xh + 2h^2 + 5h

. Dividing this expression term-by-term by

h

yields

4x + 2h + 5

.

Step-by-Step Solution

1

Evaluate

f(x+h)

for the function

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

.

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

To apply first principles, we must replace every instance of

x

with

(x+h)

and expand completely.

2

Form the numerator of the difference quotient,

f(x+h) - f(x)

.

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

Subtracting

f(x)

cancels out the terms

2x^2

and

5x

that do not contain

h

.

3

Divide the result by

h

to get the difference quotient.

\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5

Factoring out

h

from the numerator and canceling it with the denominator simplifies the quotient.

Key Concept

Differentiation from First Principles (Difference Quotient)

Estimated Time:1m 30s

Question 14Question →

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

x + 5y - 17 = 0

5x - y - 7 = 0

x - 5y + 13 = 0

x + 5y - 13 = 0

Show answer & explanation

Answer: $x + 5y - 17 = 0$

Answer

x + 5y - 17 = 0

Evaluating

y = 2x^2 - 3x + 1

at

x = 2

gives the point

(2, 3)

. Differentiating gives

\frac{dy}{dx} = 4x - 3

, which evaluates to

5

at

x = 2

. The normal line gradient is therefore

-\frac{1}{5}

. Using the point-slope formula

y - 3 = -\frac{1}{5}(x - 2)

and rearranging to standard linear form results in

x + 5y - 17 = 0

.

Step-by-Step Solution

1

Find the

y

-coordinate of the point of contact on the curve when

x = 2

.

Substitute

x = 2

into

y = 2x^2 - 3x + 1

:

y = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3

. The point is

(2, 3)

.

The line of normal passes through the point of tangency on the curve.

2

Differentiate

y = 2x^2 - 3x + 1

with respect to

x

to find the gradient of the tangent.

\frac{dy}{dx} = 4x - 3

. At

x = 2

,

m_t = 4(2) - 3 = 5

.

The derivative of a curve evaluated at a point yields the gradient of the tangent line at that point.

3

Determine the gradient of the normal line.

Since the normal is perpendicular to the tangent,

m_n = -\frac{1}{m_t} = -\frac{1}{5}

.

Perpendicular lines have gradients whose product is

-1

(

m_t \cdot m_n = -1

).

4

Form the equation of the normal line passing through

(2, 3)

with gradient

m_n = -\frac{1}{5}

.

y - 3 = -\frac{1}{5}(x - 2) \implies 5(y - 3) = -(x - 2) \implies 5y - 15 = -x + 2 \implies x + 5y - 17 = 0

.

Applying the point-slope form

y - y_1 = m(x - x_1)

establishes the linear equation.

Key Concept

Tangents and Normals to Curves

Question 15Question →

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

Show answer & explanation

Answer: -75

Answer

The value of

\frac{dy}{dx}

at

x = 2

is

-75

.

Applying the Quotient Rule

\frac{u'v - uv'}{v^2}

along with the Chain Rule to differentiate

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

yields

u'(x) = 6x(x^2+1)^2

. Evaluating at

x=2

gives

u(2)=125

,

u'(2)=300

,

v(2)=1

, and

v'(2)=3

, leading to

\frac{300(1) - 125(3)}{1^2} = -75

.

Step-by-Step Solution

1

Set up the Quotient Rule framework

Let

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

and

v(x) = 3x - 5

, so that

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

and

\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(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

x^2+1

to be multiplied.

3

Differentiate the denominator

v'(x) = 3

.

The derivative of a linear function

3x - 5

with respect to

x

is its coefficient 3.

4

Evaluate

u(x)

,

u'(x)

,

v(x)

, and

v'(x)

at

x = 2

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

,

v'(2) = 3

.

Substituting

x = 2

into each evaluated component simplifies the numerical calculation.

5

Substitute numerical values into the Quotient Rule formula

\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 16Question →

A cubic curve is defined by the equation $y = ax^3 + bx^2 + cx + d$ . The curve has a point of inflexion at $(0, 1)$ and a stationary point at $(1, 5)$ . What is the local minimum value of the function?

-3

-7

5

-1

Show answer & explanation

Answer: $-3$

Answer

The local minimum value of the function is

-3

.

By applying the conditions for a point of inflexion (

y''=0

) and a stationary point (

y'=0

), the curve equation is identified as

y = -2x^3 + 6x + 1

. Solving

y' = 0

gives

x = -1

for the local minimum, resulting in a minimum value of

-3

.

Step-by-Step Solution

1

Use the point of inflexion and given point

(0, 1)

to find

b

and

d

.

d = 1

and

b = 0

.

Since

(0,1)

lies on the curve, substituting

x=0, y=1

gives

d = 1

. The second derivative is

y'' = 6ax + 2b

. At a point of inflexion

x = 0

,

y'' = 0

, giving

2b = 0 \implies b = 0

.

2

Use the stationary point condition at

(1, 5)

to determine

a

and

c

.

a = -2

and

c = 6

.

The equation reduces to

y = ax^3 + cx + 1

. Since

(1,5)

is on the curve,

a + c + 1 = 5 \implies a + c = 4

. Also,

y' = 3ax^2 + c = 0

at

x = 1

, so

3a + c = 0

. Solving

3a + c = 0

and

a + c = 4

simultaneously yields

a = -2

and

c = 6

.

3

Find all stationary points of

y = -2x^3 + 6x + 1

and test their nature using the second derivative.

Stationary points are at

x = 1

(maximum) and

x = -1

(minimum).

Setting

y' = -6x^2 + 6 = 0

yields

x^2 = 1 \implies x = \pm 1

. The second derivative is

y'' = -12x

. For

x = -1

,

y'' = 12 > 0

, confirming a local minimum.

4

Evaluate the function at

x = -1

to find the local minimum value.

y(-1) = -3

.

Substituting

x = -1

into

y = -2x^3 + 6x + 1

yields

y = -2(-1)^3 + 6(-1) + 1 = 2 - 6 + 1 = -3

.

Key Concept

Determining curve constants using stationary points and points of inflexion, followed by identifying local extrema.

Estimated Time:3m 0s

Question 17Question →

What is the indefinite integral $\int \left( 6x^2 - \frac{4}{x^2} + 4\sin(2x)\cos(2x) \right) dx$ ?

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x)

2x^3 - \frac{4}{x} + \frac{1}{2}\cos(4x) + C

12x + \frac{8}{x^3} + 8\cos(4x) + C

Show answer & explanation

Answer: $2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C$

Answer

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

Simplifying

4\sin(2x)\cos(2x)

to

2\sin(4x)

using double-angle identities and integrating

6x^2 - 4x^{-2} + 2\sin(4x)

term-by-term yields

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

.

Step-by-Step Solution

1

Use the double-angle trigonometric identity to simplify the product term.

4\sin(2x)\cos(2x) = 2(2\sin(2x)\cos(2x)) = 2\sin(4x)

Applying the double-angle identity

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

transforms the product into a standard single sine term.

2

Express the reciprocal power with a negative exponent.

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

Rewriting the fraction in index form allows the power rule of integration to be applied directly.

3

Integrate each component term using standard integration formulas.

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

,

\int -4x^{-2} dx = \frac{-4x^{-1}}{-1} = \frac{4}{x}

, and

\int 2\sin(4x) dx = 2\left(-\frac{1}{4}\cos(4x)\right) = -\frac{1}{2}\cos(4x)

Integrating

x^n

gives

\frac{x^{n+1}}{n+1}

and integrating

\sin(ax)

yields

-\frac{1}{a}\cos(ax)

.

4

Combine the integrated terms and append the constant of integration.

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

Indefinite integrals require an arbitrary constant

C

to represent the entire family of antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 18Question →

If $y = \ln\left(\frac{1 + \sin x}{1 - \sin x}\right)$ , what is $\frac{dy}{dx}$ ?

2\sec x

-2\sec x

2\tan x

\sec x

Show answer & explanation

Answer: $2\sec x$

Answer

The derivative

\frac{dy}{dx}

is equal to

2\sec x

.

Rewriting the function as

y = \ln(1 + \sin x) - \ln(1 - \sin x)

and differentiating both terms yields

\frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}

. Combining these over the common denominator

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

simplifies to

\frac{2\cos x}{\cos^2 x} = 2\sec x

.

Step-by-Step Solution

1

Apply the logarithmic law

\ln\left(\frac{u}{v}\right) = \ln u - \ln v

y = \ln(1 + \sin x) - \ln(1 - \sin x)

Simplifies the quotient into separate terms prior to differentiation.

2

Differentiate each logarithmic term with respect to

x

using the chain rule

\frac{dy}{dx} = \frac{1}{1 + \sin x} \cdot \cos x - \frac{1}{1 - \sin x} \cdot (-\cos x) = \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}

The derivative of

\ln(f(x))

is

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

.

3

Combine the fractions over a common denominator

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

Adding two fractions with denominators

(1 + \sin x)

and

(1 - \sin x)

.

4

Use the Pythagorean trigonometric identity

1 - \sin^2 x = \cos^2 x

and simplify

\frac{dy}{dx} = \frac{2\cos x}{\cos^2 x} = \frac{2}{\cos x} = 2\sec x

Simplifying

\frac{2\cos x}{\cos^2 x}

yields

2\sec x

.

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Properties

Alternative Method

Alternatively, express

y = \ln(\sec x + \tan x)^2 = 2 \ln(\sec x + \tan x)

. The derivative of

\ln(\sec x + \tan x)

is

\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \sec x

. Multiplying by 2 gives

2\sec x

.

Estimated Time:2m 0s

Question 19Question →

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

Show answer & explanation

Answer: 22

Answer

The numerical value of the derivative at

x = 2

is 22.

Applying both the Product Rule and Chain Rule correctly yields

\frac{dy}{dx} = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}

. Evaluating this expression at

x = 2

gives

16 + 6 = 22

.

Step-by-Step Solution

1

Decompose the function into a product of two functions

Let

u(x) = x^2

and

v(x) = \sqrt{3x^2 + 4} = (3x^2 + 4)^{1/2}

.

The function

y

is expressed as the product of

u(x)

and

v(x)

, requiring the Product Rule for differentiation.

2

Find the derivative of the inner square root function using the Chain Rule

v'(x) = \frac{1}{2}(3x^2 + 4)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{3x^2 + 4}}

.

The Chain Rule states that

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

.

3

Combine derivatives using the Product Rule formula

\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

\frac{d}{dx}[u \cdot v] = u'v + uv'

.

4

Evaluate the derivative at

x = 2

\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 = 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 20Question →

Using differentiation from first principles, what is the numerical value of the derivative of the function $f(x) = 3x^2 - 4x + 1$ at $x = 2$ ?

Show answer & explanation

Answer: 8

Answer

The numerical value of the derivative of

f(x) = 3x^2 - 4x + 1

at

x = 2

is 8.

Using the first-principles limit definition, the increment

f(2+h) - f(2)

simplifies to

8h + 3h^2

. Dividing by

h

yields

8 + 3h

, which evaluates to 8 as

h \to 0

.

Step-by-Step Solution

1

Calculate

f(2)

f(2) = 5

Substitute

x = 2

into

f(x) = 3x^2 - 4x + 1

.

2

Expand

f(2+h)

f(2+h) = 5 + 8h + 3h^2

Substitute

x = 2+h

into

f(x)

and expand algebraically.

3

Simplify the difference quotient

\frac{f(2+h) - f(2)}{h}

\frac{8h + 3h^2}{h} = 8 + 3h

Subtract

f(2)

from

f(2+h)

and divide every term by

h

.

4

Evaluate the limit as

h \to 0

f'(2) = 8

As

h

approaches 0, the term

3h

vanishes, leaving 8.

Key Concept

Differentiation from First Principles

Estimated Time:1m 30s

Calculus

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