Differentiation | Jamb Mathematics

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Gear up for your differentiation exam with precision and strategy! Master the rules, from basic derivatives to advanced applications, ensuring you're ready to tackle any problem with confidence. Stay sharp, practice consistently, and approach each question with a clear, systematic method—success is in the details!

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Differentiation? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Differentiation together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problem on limit of a function

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Question 1:

Evaluate

\lim_{x \to 3} (2x + 5)

Solution:

Since the function

f(x) = 2x + 5

is continuous, we substitute

x = 3

directly:

\lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11

Question 2:

Evaluate

\lim_{x \to 4} (x^2 - 2x + 1)

Solution:

The function is a polynomial, so we directly substitute

x = 4

\lim_{x \to 4} (4^2 - 2(4) + 1) = 16 - 8 + 1 = 9

Question 3:

Evaluate

\lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Solution:

Substituting

x = 2

gives

\frac{4 - 4}{2 - 2} = \frac{0}{0}

, an indeterminate form.
Factor the numerator:

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

.
Cancel

(x - 2)

\lim_{x \to 2} (x + 2) = 2 + 2 = 4

Question 4:

Find

\lim_{x \to 0} \frac{\sin x}{x}

Solution:

This is a standard limit result:

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

Question 5:

Evaluate

\lim_{x \to \infty} \frac{3x^2 + 5}{7x^2 + 2}

Solution:

Divide numerator and denominator by

x^2

\lim_{x \to \infty} \frac{3 + \frac{5}{x^2}}{7 + \frac{2}{x^2}} = \frac{3}{7}

Question 6:

Find

\lim_{x \to -\infty} \frac{2x^3 + 5}{4x^3 - 3x}

Solution:

Divide by

x^3

\lim_{x \to -\infty} \frac{2 + \frac{5}{x^3}}{4 - \frac{3}{x^2}} = \frac{2}{4} = \frac{1}{2}

Question 7:

Find

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

Solution:

Factor numerator as

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

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

Question 8:

Evaluate

\lim_{x \to 0} \frac{1 - \cos x}{x^2}

Solution:

Using

\cos x \approx 1 - \frac{x^2}{2}

for small

x

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

.
Thus,

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Question 9:

Find

\lim_{x \to 0} \frac{e^x - 1}{x}

Solution:

Using

e^x \approx 1 + x

for small

x

\frac{(1 + x) - 1}{x} = \frac{x}{x} = 1

.
Thus,

\lim_{x \to 0} \frac{e^x - 1}{x} = 1

Question 10:

Evaluate

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Solution:

Factor numerator:

\frac{(x - 3)(x + 3)}{x - 3}

.
Cancel

(x - 3)

\lim_{x \to 3} (x + 3) = 3 + 3 = 6

Question 11:

Find

\lim_{x \to 0} \frac{\tan x}{x}

Solution:

Using the standard limit result:

\lim_{x \to 0} \frac{\tan x}{x} = 1

Question 12:

Find

\lim_{x \to 1} \frac{x^4 - 1}{x - 1}

Solution:

Factor numerator:

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

.
Cancel

(x - 1)

\lim_{x \to 1} (x^3 + x^2 + x + 1) = 1 + 1 + 1 + 1 = 4

Question 13:

Find

\lim_{x \to 0} \frac{x - \sin x}{x^3}

Solution:

Using the approximation

\sin x \approx x - \frac{x^3}{6}

\frac{x - (x - \frac{x^3}{6})}{x^3} = \frac{x^3/6}{x^3} = \frac{1}{6}

.
Thus,

\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}

Question 14:

Evaluate

\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}

Solution:

Factor numerator:

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

.
Factor denominator:

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

.
Cancel

(x - 2)

\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{4 + 4 + 4}{2 + 2} = \frac{12}{4} = 3

Question 15:

Evaluate

\lim_{x \to 0} \frac{\ln(1 + x)}{x}

Solution:

Using the standard limit result:

\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

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Calculation problem on Differentiation of explicit algebraic expression

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Question 1:

Find

\frac{d}{dx} (3x^2 + 5x - 7)

Solution:

Differentiate term by term:

\frac{d}{dx} (3x^2) = 6x

\frac{d}{dx} (5x) = 5

\frac{d}{dx} (-7) = 0

Thus,

\frac{d}{dx} (3x^2 + 5x - 7) = 6x + 5

Question 2:

Find

\frac{d}{dx} (x^3 - 4x + 7)

Solution:

Using power rule:

\frac{d}{dx} (x^3) = 3x^2

\frac{d}{dx} (-4x) = -4

\frac{d}{dx} (7) = 0

Thus,

\frac{d}{dx} (x^3 - 4x + 7) = 3x^2 - 4

Question 3:

Find

\frac{d}{dx} (x^4 + 2x^2 - x + 5)

Solution:

Using power rule:

\frac{d}{dx} (x^4) = 4x^3

\frac{d}{dx} (2x^2) = 4x

\frac{d}{dx} (-x) = -1

\frac{d}{dx} (5) = 0

Thus,

\frac{d}{dx} (x^4 + 2x^2 - x + 5) = 4x^3 + 4x - 1

Question 4:

Find

\frac{d}{dx} (5x^5 - 3x^3 + 7)

Solution:

Differentiate each term:

\frac{d}{dx} (5x^5) = 25x^4

\frac{d}{dx} (-3x^3) = -9x^2

\frac{d}{dx} (7) = 0

Thus,

\frac{d}{dx} (5x^5 - 3x^3 + 7) = 25x^4 - 9x^2

Question 5:

Find

\frac{d}{dx} (x^6 - 2x^4 + x - 8)

Solution:

\frac{d}{dx} (x^6) = 6x^5

\frac{d}{dx} (-2x^4) = -8x^3

\frac{d}{dx} (x) = 1

\frac{d}{dx} (-8) = 0

Thus,

\frac{d}{dx} (x^6 - 2x^4 + x - 8) = 6x^5 - 8x^3 + 1

Question 6:

Find

\frac{d}{dx} (7x^2 + 4x - 3)

Solution:

\frac{d}{dx} (7x^2) = 14x

\frac{d}{dx} (4x) = 4

\frac{d}{dx} (-3) = 0

Thus,

\frac{d}{dx} (7x^2 + 4x - 3) = 14x + 4

Question 7:

Find

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

Solution:

\frac{d}{dx} (x^3) = 3x^2

\frac{d}{dx} (x^2) = 2x

\frac{d}{dx} (x) = 1

\frac{d}{dx} (1) = 0

Thus,

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

Question 8:

Find

\frac{d}{dx} (x^5 - x^3 + 2x)

Solution:

\frac{d}{dx} (x^5) = 5x^4

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

\frac{d}{dx} (2x) = 2

Thus,

\frac{d}{dx} (x^5 - x^3 + 2x) = 5x^4 - 3x^2 + 2

Question 9:

Find

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

Solution:

Using the power rule:

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

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

Thus,

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

Question 10:

Find

\frac{d}{dx} (x^4 + 5x^{\frac{1}{3}})

Solution:

Using the power rule:

\frac{d}{dx} (x^4) = 4x^3

\frac{d}{dx} (5x^{\frac{1}{3}}) = \frac{5}{3}x^{-\frac{2}{3}}

Thus,

\frac{d}{dx} (x^4 + 5x^{\frac{1}{3}}) = 4x^3 + \frac{5}{3}x^{-\frac{2}{3}}

Problem 11: Differentiate implicitly

x^2 + y^2 = \sin(y)

Solution:

Differentiate both sides with respect to $x$ , treating $y$ as an implicit function of $x$ . $2x + 2y \frac{dy}{dx} = \cos(y) \frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} (2y - \cos y) = -2x$ $\frac{dy}{dx} = \frac{-2x}{2y - \cos y}$

Problem 12: Differentiate implicitly

\tan(y) + x^3 = y^3

Solution:

Differentiate both sides: $\sec^2 y \frac{dy}{dx} + 3x^2 = 3y^2 \frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} (\sec^2 y - 3y^2) = -3x^2$ $\frac{dy}{dx} = \frac{-3x^2}{\sec^2 y - 3y^2}$

Seconnd Derivative Problems

Problem 13: Find

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

for

y = \sin(2x)

Solution:

First derivative: $\frac{dy}{dx} = 2\cos(2x)$
Second derivative: $\frac{d^2y}{dx^2} = -4\sin(2x)$

Problem 14: Find

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

for

y = \tan(x)

Solution:

First derivative: $\frac{dy}{dx} = \sec^2 x$
Second derivative: $\frac{d^2y}{dx^2} = 2\sec^2 x \tan x$

Inverse Trigonometric Function Problems

Problem 15: Differentiate

y = \arcsin(x^2)

Solution:

Use the chain rule: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x$
Simplify: $\frac{dy}{dx} = \frac{2x}{\sqrt{1 - x^4}}$

Problem 16: Differentiate

y = \arctan(3x)

Solution:

Use the chain rule: $\frac{dy}{dx} = \frac{1}{1 + (3x)^2} \cdot 3$
Simplify: $\frac{dy}{dx} = \frac{3}{1 + 9x^2}$

Higher Order Differentiation Problems

Problem 17: Find

\frac{d^3y}{dx^3}

for

y = \cos(4x)

Solution:

First derivative: $\frac{dy}{dx} = -4\sin(4x)$
Second derivative: $\frac{d^2y}{dx^2} = -16\cos(4x)$
Third derivative: $\frac{d^3y}{dx^3} = 64\sin(4x)$

Problem 18: Find

\frac{d^3y}{dx^3}

for

y = e^x \cos x

Solution:

First derivative: $\frac{dy}{dx} = e^x \cos x - e^x \sin x$
Second derivative: $\frac{d^2y}{dx^2} = e^x \cos x - e^x \sin x - e^x \sin x - e^x \cos x$ $\frac{d^2y}{dx^2} = -2e^x \sin x$
Third derivative: $\frac{d^3y}{dx^3} = -2e^x \cos x - 2e^x \sin x$ $\frac{d^3y}{dx^3} = -2e^x (\cos x + \sin x)$

Mixed Application Problems

Problem 19: Differentiate

y = x^2 \arcsin x

Solution:

Use the product rule:
- First term: $u = x^2 \Rightarrow u' = 2x$
- Second term: $v = \arcsin x \Rightarrow v' = \frac{1}{\sqrt{1 - x^2}}$
Apply the product rule: $\frac{dy}{dx} = 2x \arcsin x + \frac{x^2}{\sqrt{1 - x^2}}$

Problem 20: Differentiate

y = \frac{\tan x}{x}

Solution:

Use the quotient rule:
- $u = \tan x \Rightarrow u' = \sec^2 x$
- $v = x \Rightarrow v' = 1$
Apply the quotient rule: $\frac{dy}{dx} = \frac{(x\sec^2 x - \tan x)}{x^2}$

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Differentiation | Jamb Mathematics

Calculation problem on limit of a function

Question 1:

Solution:

Question 2:

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Question 3:

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Question 4:

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Question 10:

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Question 11:

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Question 12:

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Question 14:

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Question 15:

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Calculation problem on Differentiation of explicit algebraic expression

Question 1:

Solution:

Question 2:

Solution:

Question 3:

Solution:

Question 4:

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Question 5:

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Question 6:

Solution:

Question 7:

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Question 8:

Solution:

Question 9:

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Question 10:

Solution:

Problem 11: Differentiate implicitly

Problem 12: Differentiate implicitly

Seconnd Derivative Problems

Problem 13: Find

Problem 14: Find

Inverse Trigonometric Function Problems

Problem 15: Differentiate

Problem 16: Differentiate

Higher Order Differentiation Problems

Problem 17: Find

Problem 18: Find

Mixed Application Problems

Problem 19: Differentiate

Problem 20: Differentiate

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