Question

derivatives of logarithmic functions: problem 3 (1 point)
use logarithmic differentiation to find the derivative of the function.
$y = sqrt{x}e^{x^{2}}(x^{2}+2)^{10}$
$y=square$

Explanation:

Step1: Take natural - log of both sides

Take the natural logarithm of \(y = \sqrt{x}e^{x^{2}}(x^{2}+2)^{10}\). Using the properties of logarithms \(\ln(ab)=\ln a+\ln b\) and \(\ln(a^{n}) = n\ln a\), we have \(\ln y=\ln(\sqrt{x})+\ln(e^{x^{2}})+\ln((x^{2}+2)^{10})\). Since \(\sqrt{x}=x^{\frac{1}{2}}\) and \(\ln(e^{x^{2}})=x^{2}\), \(\ln y=\frac{1}{2}\ln x + x^{2}+10\ln(x^{2}+2)\).

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 \(\frac{1}{2}\ln x\) is \(\frac{1}{2x}\), the derivative of \(x^{2}\) is \(2x\), and the derivative of \(10\ln(x^{2}+2)\) using the chain - rule is \(\frac{10\times2x}{x^{2}+2}=\frac{20x}{x^{2}+2}\). So \(\frac{y'}{y}=\frac{1}{2x}+2x+\frac{20x}{x^{2}+2}\).

Step3: Solve for \(y'\)

Multiply both sides by \(y=\sqrt{x}e^{x^{2}}(x^{2}+2)^{10}\). Then \(y'=\sqrt{x}e^{x^{2}}(x^{2}+2)^{10}(\frac{1}{2x}+2x+\frac{20x}{x^{2}+2})\).

Answer:

\(y'=\sqrt{x}e^{x^{2}}(x^{2}+2)^{10}(\frac{1}{2x}+2x+\frac{20x}{x^{2}+2})\)