Question

find dy/dx if y = sec(8x²)

Explanation:

Step1: Identify the outer and inner functions

The function \( y = \sec(8x^2) \) is a composite function. Let \( u = 8x^2 \) (inner function) and \( y=\sec(u) \) (outer function).

Step2: Differentiate the outer function

The derivative of \( \sec(u) \) with respect to \( u \) is \( \sec(u)\tan(u) \). So, \( \frac{dy}{du}=\sec(u)\tan(u) \).

Step3: Differentiate the inner function

The derivative of \( u = 8x^2 \) with respect to \( x \) is \( \frac{du}{dx}=16x \) (using the power rule: \( \frac{d}{dx}(ax^n)=nax^{n - 1} \), here \( a = 8 \), \( n = 2 \), so \( 2\times8x^{2-1}=16x \)).

Step4: Apply the chain rule

The chain rule states that \( \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \). Substitute \( \frac{dy}{du}=\sec(u)\tan(u) \) and \( \frac{du}{dx}=16x \), and replace \( u \) with \( 8x^2 \). So, \( \frac{dy}{dx}=\sec(8x^2)\tan(8x^2)\times16x = 16x\sec(8x^2)\tan(8x^2) \).

Answer:

\( 16x\sec(8x^2)\tan(8x^2) \)