Question

use logarithmic differentiation to find the derivative of the function. y = x^7 sin(x) y(x) =

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(x^{7}\sin(x))$
Using the property $\ln(ab)=\ln a+\ln b$, we get $\ln y = 7\ln x+\ln(\sin x)$

Step2: Differentiate both sides with respect to $x$

The derivative of $\ln y$ with respect to $x$ is $\frac{y'}{y}$ by the chain - rule.
The derivative of $7\ln x$ with respect to $x$ is $\frac{7}{x}$, and the derivative of $\ln(\sin x)$ with respect to $x$ is $\frac{\cos x}{\sin x}=\cot x$ using the chain - rule.
So, $\frac{y'}{y}=\frac{7}{x}+\cot x$

Step3: Solve for $y'$

Multiply both sides by $y$. Since $y = x^{7}\sin(x)$, we have $y'=y(\frac{7}{x}+\cot x)$
Substitute $y$ back in: $y'=x^{7}\sin(x)(\frac{7}{x}+\cot x)=7x^{6}\sin(x)+x^{7}\cos(x)$

Answer:

$7x^{6}\sin(x)+x^{7}\cos(x)$