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

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

Cevap: 34

The value of

\frac{dy}{dx}

x = 2

34

Applying the product rule

\frac{dy}{dx} = u'v + uv'

alongside the chain rule gives

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

. Evaluating at

x = 2

yields

4(1) + 30(1) = 34

Identify component functions for the product rule

Let

u(x) = x^2 + 1

and

v(x) = (2x - 3)^3

The function

y

is a product of two differentiable functions.

Differentiate each component function

u'(x) = 2x

and

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

The power rule gives

u'(x)

and the chain rule gives

v'(x)

by multiplying by the derivative of the inner function

(2x - 3)

Apply the product rule formula

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

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

To find the general derivative of a product of functions.

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34

To calculate the specific numerical value of the derivative at

x = 2

Product and Chain Rules of Differentiation

Tahmini Süre:1m 30s

Soru