use logarithmic differentiation to find the derivative of y with respect to x.\ny = x^{8x +…

use logarithmic differentiation to find the derivative of y with respect to x.\ny = x^{8x + 7}\n\\frac{dy}{dx}=\\square
Answer
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:
$x^{8x + 7}(8\ln x+\frac{8x + 7}{x})$