use logarithmic differentiation to find the derivative of the function. y = ((sin(x))^6 (tan(x))^4)/((x^2 +…

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\left(\frac{(\sin x)^{6}(\tan x)^{4}}{(x^{2}+2)^{2}}\right)$ Using the properties of logarithms $\ln\left(\frac{a}{b}\right)=\ln a-\ln b$ and $\ln(ab)=\ln a+\ln b$, we get $\ln y = 6\ln(\sin x)+4\ln(\tan x)-2\ln(x^{2}+2)$.

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 $6\ln(\sin x)$ with respect to $x$ is $6\cdot\frac{\cos x}{\sin x}=6\cot x$ (using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u'}{u}$, where $u = \sin x$ and $u'=\cos x$). The derivative of $4\ln(\tan x)$ with respect to $x$ is $4\cdot\frac{\sec^{2}x}{\tan x}=\frac{4}{\cos^{2}x}\cdot\frac{\cos x}{\sin x}=\frac{4}{\sin x\cos x}=4\csc x\sec x$. The derivative of $-2\ln(x^{2}+2)$ with respect to $x$ is $-2\cdot\frac{2x}{x^{2}+2}=-\frac{4x}{x^{2}+2}$. So, $\frac{y'}{y}=6\cot x + 4\csc x\sec x-\frac{4x}{x^{2}+2}$.

Step3: Solve for $y'$

Multiply both sides by $y$: $y'=y\left(6\cot x + 4\csc x\sec x-\frac{4x}{x^{2}+2}\right)$. Substitute $y = \frac{(\sin x)^{6}(\tan x)^{4}}{(x^{2}+2)^{2}}$ back into the equation: $y'=\frac{(\sin x)^{6}(\tan x)^{4}}{(x^{2}+2)^{2}}\left(6\cot x + 4\csc x\sec x-\frac{4x}{x^{2}+2}\right)$.

Answer:

$\frac{(\sin x)^{6}(\tan x)^{4}}{(x^{2}+2)^{2}}\left(6\cot x + 4\csc x\sec x-\frac{4x}{x^{2}+2}\right)$