use logarithmic differentiation to determine y for the equation y = (cos(3x))^4x. answer o y =…

Explanation:

Step1: Take natural - log of both sides

Take the natural - log of $y = (\cos(3x))^{4x}$. So, $\ln y=\ln((\cos(3x))^{4x})$. Using the power - rule of logarithms $\ln a^b = b\ln a$, we get $\ln y = 4x\ln(\cos(3x))$.

Step2: Differentiate both sides with respect to $x$

Differentiate the left - hand side: $\frac{d}{dx}(\ln y)=\frac{1}{y}y'$. Differentiate the right - hand side using the product rule $(uv)' = u'v+uv'$, where $u = 4x$ and $v=\ln(\cos(3x))$. $u' = 4$. To find $v'$, use the chain rule. Let $u=\cos(3x)$, so $v=\ln u$. Then $\frac{dv}{du}=\frac{1}{u}$ and $\frac{du}{dx}=- 3\sin(3x)$. So, $v'=\frac{-3\sin(3x)}{\cos(3x)}=-3\tan(3x)$. By the product rule, $\frac{d}{dx}(4x\ln(\cos(3x)))=4\ln(\cos(3x))+4x(-3\tan(3x))=4\ln(\cos(3x))-12x\tan(3x)$.

Step3: Solve for $y'$

Since $\frac{1}{y}y' = 4\ln(\cos(3x))-12x\tan(3x)$ and $y = (\cos(3x))^{4x}$, then $y'=(\cos(3x))^{4x}(4\ln(\cos(3x))-12x\tan(3x))$.

Answer:

$y' = (\cos(3x))^{4x}(4\ln(\cos(3x))-12x\tan(3x))$