Test Index

JAMB Mathematics 2024 Paper

© africaexams.com

Question : 19

Total: 50

Differentiate y = (5x + 1)

^{4}

20 (5x + 1) $^{3}$
4(5x + 1) $^{3}$
5(5x + 1) $^{3}$
(5x) $^{3}$

Solution:

Concept:

Use the chain rule: differentiate the outer power first, then multiply by the derivative of the inner linear function.

Explanation:

Let

u = 5x + 1

.

Then

\frac{du}{dx} = 5

.

The function becomes

y = u^4

.

Differentiate with respect to

u

:

\frac{dy}{du} = 4u^3

.

Apply the chain rule:

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

.

So

\frac{dy}{dx} = 4u^3 \times 5 = 20u^3

.

Substitute back

u = 5x + 1

to get

\frac{dy}{dx} = 20(5x + 1)^3

.

Answer:

A.

20(5x + 1)^3

© africaexams.com

Go to Question:

Prev Question

Next Question