use logarithmic differentiation to find the derivative of the function.\n(y=(x^{3}+2)^{2}(x^{4}+4)^{4})\n(y(x…

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\left[(x^{3}+2)^{2}(x^{4}+4)^{4}\right]$ Using the property $\ln(ab)=\ln a+\ln b$, we get $\ln y = 2\ln(x^{3}+2)+4\ln(x^{4}+4)$.

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 $2\ln(x^{3}+2)$ with respect to $x$ is $2\times\frac{3x^{2}}{x^{3}+2}=\frac{6x^{2}}{x^{3}+2}$ (using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u'}{u}$, where $u = x^{3}+2$ and $u'=3x^{2}$). The derivative of $4\ln(x^{4}+4)$ with respect to $x$ is $4\times\frac{4x^{3}}{x^{4}+4}=\frac{16x^{3}}{x^{4}+4}$ (using the chain - rule with $u = x^{4}+4$ and $u' = 4x^{3}$). So, $\frac{y'}{y}=\frac{6x^{2}}{x^{3}+2}+\frac{16x^{3}}{x^{4}+4}$.

Step3: Solve for $y'$

Multiply both sides by $y=(x^{3}+2)^{2}(x^{4}+4)^{4}$: $y'=(x^{3}+2)^{2}(x^{4}+4)^{4}\left(\frac{6x^{2}}{x^{3}+2}+\frac{16x^{3}}{x^{4}+4}\right)$ $y'=(x^{3}+2)(x^{4}+4)^{4}\times6x^{2}+(x^{3}+2)^{2}(x^{4}+4)^{3}\times16x^{3}$ $y' = 6x^{2}(x^{3}+2)(x^{4}+4)^{4}+16x^{3}(x^{3}+2)^{2}(x^{4}+4)^{3}$

Answer:

$6x^{2}(x^{3}+2)(x^{4}+4)^{4}+16x^{3}(x^{3}+2)^{2}(x^{4}+4)^{3}$