If y = x^3(2x - 1)^4, find the value of \frac{dy}{dx} at x = 1. — JAMB UTME

If $y = x^3(2x - 1)^4$ , find the value of $\frac{dy}{dx}$ at $x = 1$ .

Answer: 11

The numerical value of

\frac{dy}{dx}

x = 1

11

Applying the product rule

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

along with the chain rule for the expression

(2x - 1)^4

gives

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

. Evaluating this at

x = 1

gives

8(1)^3(1)^3 + 3(1)^2(1)^4 = 8 + 3 = 11

Set up the product rule for

y = u \cdot v

, where

u = x^3

and

v = (2x - 1)^4

u = x^3

and

v = (2x - 1)^4

The function is expressed as the product of two algebraic terms.

Find the derivative of each function component.

\frac{du}{dx} = 3x^2

and

\frac{dv}{dx} = 8(2x - 1)^3

The power rule gives

\frac{du}{dx} = 3x^2

, and applying the chain rule to

(2x - 1)^4

yields

4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3

Substitute components into the product rule formula

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

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

Combining

u\frac{dv}{dx}

and

v\frac{du}{dx}

provides the full expression for the derivative.

Evaluate the derivative at

x = 1

\frac{dy}{dx}\Big|_{x=1} = 8(1)^3(2(1) - 1)^3 + 3(1)^2(2(1) - 1)^4 = 8 + 3 = 11

Substituting

x = 1

simplifies the terms to

8(1) + 3(1) = 11

Rules of Differentiation (Product and Chain Rules)

Question