Question

differentiate the function.

$y = \left(4x^2 - 14\
ight)^{-12}$

$\dfrac{dy}{dx} = \square$

Explanation:

Step1: Apply Chain Rule

Let \( u = 4x^2 - 14 \), so \( y = u^{-12} \). The chain rule states \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \). First, find \( \frac{dy}{du} \):
\( \frac{dy}{du}=-12u^{-13} \)

Step2: Find \( \frac{du}{dx} \)

Differentiate \( u = 4x^2 - 14 \) with respect to \( x \):
\( \frac{du}{dx}=8x \)

Step3: Substitute Back

Substitute \( u = 4x^2 - 14 \) and the derivatives into the chain rule:
\( \frac{dy}{dx}=-12(4x^2 - 14)^{-13}\cdot8x \)
Simplify the coefficient: \( -12\times8x=-96x \)
So \( \frac{dy}{dx}=-96x(4x^2 - 14)^{-13} \) or \( \frac{dy}{dx}=\frac{-96x}{(4x^2 - 14)^{13}} \)

Answer:

\( -96x(4x^2 - 14)^{-13} \) (or equivalent form \( \frac{-96x}{(4x^2 - 14)^{13}} \))