Question

use logarithmic differentiation to find the derivative of y with respect to x.
y = x^{8x + 7}
\frac{dy}{dx}=\square

Explanation:

Step1: Take natural - log of both sides

$\ln y=(8x + 7)\ln x$

Step2: Differentiate both sides with respect to x

Using the product rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = 8x + 7$ and $v=\ln x$.
The derivative of the left - hand side is $\frac{1}{y}\frac{dy}{dx}$, and the derivative of the right - hand side:
$u^\prime=8$ and $v^\prime=\frac{1}{x}$, so $(8x + 7)^\prime\ln x+(8x + 7)(\ln x)^\prime=8\ln x+\frac{8x + 7}{x}$
So, $\frac{1}{y}\frac{dy}{dx}=8\ln x+\frac{8x + 7}{x}$

Step3: Solve for $\frac{dy}{dx}$

Multiply both sides by $y=x^{8x + 7}$
$\frac{dy}{dx}=x^{8x + 7}(8\ln x+\frac{8x + 7}{x})$

Answer:

$x^{8x + 7}(8\ln x+\frac{8x + 7}{x})$