use logarithmic differentiation to find the derivative of the function.\ny = (tan(x))^{5/x}\ny =

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(\tan(x))^{\frac{5}{x}}=\frac{5}{x}\ln(\tan(x))$

Step2: Differentiate both sides with respect to $x$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = \frac{5}{x}=5x^{-1}$ and $v=\ln(\tan(x))$. The derivative of $u$ with respect to $x$ is $u^\prime=- 5x^{-2}=-\frac{5}{x^{2}}$. The derivative of $v$ with respect to $x$: Let $t = \tan(x)$, then $v=\ln(t)$. By the chain - rule, $v^\prime=\frac{1}{t}\cdot\sec^{2}(x)=\frac{\sec^{2}(x)}{\tan(x)}$. So, $\frac{y^\prime}{y}=-\frac{5}{x^{2}}\ln(\tan(x))+\frac{5}{x}\cdot\frac{\sec^{2}(x)}{\tan(x)}$

Step3: Solve for $y^\prime$

Multiply both sides by $y = (\tan(x))^{\frac{5}{x}}$ $y^\prime=(\tan(x))^{\frac{5}{x}}\left(-\frac{5\ln(\tan(x))}{x^{2}}+\frac{5\sec^{2}(x)}{x\tan(x)}\right)$

Answer:

$(\tan(x))^{\frac{5}{x}}\left(-\frac{5\ln(\tan(x))}{x^{2}}+\frac{5\sec^{2}(x)}{x\tan(x)}\right)$