use logarithmic differentiation to find the derivative of the function. $y = sqrt{xe^{x^{3}}}(x^{2}+3)^{10}$

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\left(\sqrt{xe^{x^{3}}}(x^{2}+3)^{10}\right)$ Using properties of logarithms $\ln(ab)=\ln a+\ln b$ and $\ln a^{n}=n\ln a$, we get $\ln y=\frac{1}{2}(\ln x+\ln e^{x^{3}})+10\ln(x^{2}+3)=\frac{1}{2}(\ln x + x^{3})+10\ln(x^{2}+3)$.

Step2: Differentiate both sides with respect to $x$

The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}y'$ by the chain - rule. The derivative of $\frac{1}{2}(\ln x + x^{3})$ with respect to $x$ is $\frac{1}{2}\left(\frac{1}{x}+3x^{2}\right)$ and the derivative of $10\ln(x^{2}+3)$ with respect to $x$ is $\frac{10\times2x}{x^{2}+3}$. So, $\frac{1}{y}y'=\frac{1}{2}\left(\frac{1}{x}+3x^{2}\right)+\frac{20x}{x^{2}+3}$.

Step3: Solve for $y'$

Multiply both sides by $y=\sqrt{xe^{x^{3}}}(x^{2}+3)^{10}$: $y'=\sqrt{xe^{x^{3}}}(x^{2}+3)^{10}\left[\frac{1}{2}\left(\frac{1}{x}+3x^{2}\right)+\frac{20x}{x^{2}+3}\right]$.

Answer:

$y'=\sqrt{xe^{x^{3}}}(x^{2}+3)^{10}\left[\frac{1}{2}\left(\frac{1}{x}+3x^{2}\right)+\frac{20x}{x^{2}+3}\right]$