If y = (2x^2 + 1)^3, what is the numerical value of \frac{dy}{dx} evaluated at x = 1? — JAMB UTME

If $y = (2x^2 + 1)^3$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

108

Differentiating

y = (2x^2 + 1)^3

using the chain rule yields

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

. Substituting

x = 1

gives

12(1)(3)^2 = 108

Identify inner and outer functions for the composite expression

Let

u = 2x^2 + 1

, so

y = u^3

The function requires the application of the Chain Rule.

Compute the derivatives of the outer and inner functions

\frac{dy}{du} = 3u^2 = 3(2x^2 + 1)^2

and

\frac{du}{dx} = 4x

Apply the power rule to both components.

Apply the Chain Rule formula

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

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot (4x) = 12x(2x^2 + 1)^2

Multiply the outer derivative by the inner derivative.

Substitute

x = 1

into the derivative

\frac{dy}{dx}\Big|_{x=1} = 12(1)(2(1)^2 + 1)^2 = 12(3)^2 = 12 \cdot 9 = 108

Evaluate at the given x-value.

Chain Rule of Differentiation

Estimated Time:1m 0s

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