Question

use logarithmic differentiation to find the derivative of y with respect to the given independent variable.

( y = ( sin 7 x ) ^ { 4 x } )

( \frac { d y } { d x } = )
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Take natural logarithm on both sides

$$\ln y=\ln(\sin7x)^{4x}=4x\ln(\sin7x)$$

Step2: Differentiate both sides with respect to \(x\)

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = 4x\) and \(v=\ln(\sin7x)\).

First, \(u^\prime=4\).

For \(v=\ln(\sin7x)\), using the chain rule \((\ln u)^\prime=\frac{u^\prime}{u}\), let \(u = \sin7x\), then \(u^\prime=7\cos7x\). So \(v^\prime=\frac{7\cos7x}{\sin7x}=7\cot7x\).

The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y}\frac{dy}{dx}\).

Differentiating the right - hand side: \(\frac{d}{dx}(4x\ln(\sin7x))=4\ln(\sin7x)+4x\times7\cot7x=4\ln(\sin7x)+28x\cot7x\)

So \(\frac{1}{y}\frac{dy}{dx}=4\ln(\sin7x)+28x\cot7x\)

Step3: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y = (\sin7x)^{4x}\)

\(\frac{dy}{dx}=(\sin7x)^{4x}(4\ln(\sin7x)+28x\cot7x)\)

Answer:

\((\sin7x)^{4x}(4\ln(\sin7x)+28x\cot7x)\)