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

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

Answer: 32

The numerical value of

\frac{dy}{dx}

evaluated at

x = 1

32

Using the product rule together with the chain rule for composite functions, we find

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

and

v'(x) = 4x(x^2 + 1)

. Evaluating at

x = 1

gives

u(1)=1

u'(1)=6

v(1)=4

, and

v'(1)=8

. Calculating

\frac{dy}{dx} = u'(1)v(1) + u(1)v'(1) = 6(4) + 1(8) = 32

Set up the product rule for

y = u(x)v(x)

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

and

v(x) = (x^2 + 1)^2

The given function is a product of two composite expressions.

Differentiate

u(x)

using the chain rule

u'(x) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2

Differentiating the outer power function and multiplying by the derivative of the inner function

2x - 1

Differentiate

v(x)

using the chain rule

v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)

Differentiating the outer power function and multiplying by the derivative of the inner function

x^2 + 1

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

and evaluate at

x = 1

u(1) = 1

u'(1) = 6

v(1) = 4

v'(1) = 8

, giving

\frac{dy}{dx}\Big|_{x=1} = (6)(4) + (1)(8) = 32

Substituting

x = 1

into each individual term simplifies the arithmetic before combining.

Combined application of the Product Rule and Chain Rule

Question