use logarithmic differentiation to find the derivative of the function. y = (cos 9x)^x y = need help? read…

Explanation:

Step1: Take natural - log of both sides

Given $y = (\cos9x)^x$, take $\ln$ of both sides: $\ln y=x\ln(\cos9x)$.

Step2: Differentiate both sides with respect to $x$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v=\ln(\cos9x)$. The derivative of $u=x$ is $u^\prime = 1$, and for $v=\ln(\cos9x)$, using the chain - rule, if $v=\ln(u)$ and $u = \cos9x$, then $v^\prime=\frac{1}{\cos9x}\cdot(-\sin9x)\cdot9=- 9\tan9x$. So, $\frac{1}{y}y^\prime=\ln(\cos9x)+x(-9\tan9x)$.

Step3: Solve for $y^\prime$

Multiply both sides by $y = (\cos9x)^x$: $y^\prime=(\cos9x)^x(\ln(\cos9x)-9x\tan9x)$.

For $y = (\tan x)^{\frac{7}{x}}$:

Step1: Take natural - log of both sides

$\ln y=\frac{7}{x}\ln(\tan x)$.

Step2: Differentiate both sides with respect to $x$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=\frac{7}{x}=7x^{-1}$ and $v = \ln(\tan x)$. The derivative of $u$ is $u^\prime=-7x^{-2}=-\frac{7}{x^{2}}$, and for $v=\ln(\tan x)$, using the chain - rule, if $v = \ln(u)$ and $u=\tan x$, then $v^\prime=\frac{1}{\tan x}\cdot\sec^{2}x=\frac{\sec^{2}x}{\tan x}=\csc x\sec x$. So, $\frac{1}{y}y^\prime=-\frac{7}{x^{2}}\ln(\tan x)+\frac{7}{x}\csc x\sec x$.

Step3: Solve for $y^\prime$

Multiply both sides by $y = (\tan x)^{\frac{7}{x}}$: $y^\prime=(\tan x)^{\frac{7}{x}}\left(\frac{7}{x}\csc x\sec x-\frac{7}{x^{2}}\ln(\tan x)\right)$.

Answer:

For $y = (\cos9x)^x$, $y^\prime=(\cos9x)^x(\ln(\cos9x)-9x\tan9x)$; for $y = (\tan x)^{\frac{7}{x}}$, $y^\prime=(\tan x)^{\frac{7}{x}}\left(\frac{7}{x}\csc x\sec x-\frac{7}{x^{2}}\ln(\tan x)\right)$