$\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\end{align}
$\textit{thot this would help but what next??}$
$\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\end{align}
$\textit{thot this would help but what next??}$
Use implicit differentiation, knowing that $\d{}{x}\ln\left({x}\right)=\frac{1}{x}$ and $\d{}{x}\ln\left({y}\right)=\frac{1}{y}\d{y}{x}$.
$\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
Last edited by karush; January 5th, 2017 at 19:55. Reason: latex
$\displaystyle{y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}}}$
Let $f=x \, \sqrt[]{x^2+1}$ and $g=(x+1)^{2/3}$.
Then $\displaystyle{y'=\frac{f'\cdot g-f\cdot g'}{g^2}} \ \ \ (\star)$
We have the following:
\begin{align*}f' & =(x)' \, \sqrt[]{x^2+1}+x \, (\sqrt[]{x^2+1})'= \sqrt[]{x^2+1}+x \, \frac{1}{2\sqrt[]{x^2+1}}\cdot (x^2+1)' =\sqrt[]{x^2+1}+x \, \frac{2x}{2\sqrt[]{x^2+1}} \\ &=\sqrt[]{x^2+1}+ \, \frac{x^2}{\sqrt[]{x^2+1}} =\frac{\sqrt[]{x^2+1}^2+x^2}{\sqrt[]{x^2+1}}=\frac{x^2+1+x^2}{\sqrt[]{x^2+1}}\\ &=\frac{2x^2+1}{\sqrt[]{x^2+1}}\end{align*}
$$g'=\frac{2}{3}(x+1)^{2/3-1}=\frac{2}{3}(x+1)^{-1/3}$$
$$g^2=(x+1)^{4/3}$$
So, substituting these at the relation $(\star)$ we get:
\begin{align*}y' &=\frac{\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}} \\ &=\frac{\sqrt{x^2+1}\left (\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{(x+1)^{1/3}\left (\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{1/3}(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}}{(x+1)^{5/3}\sqrt{x^2+1}} \\ & =\frac{3\left (\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}\right )}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^2+1\right )\cdot (x+1)-2x \, (x^2+1)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^3+x+2x^2+1\right )-2 \, (x^3+x)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{6x^3+3x+6x^2+3-2x^3-2x}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{4x^3+6x^2+x+3}{3(x+1)^{5/3}\sqrt{x^2+1}}\end{align*}
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
$\displaystyle
y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \therefore \frac{y}{x}=\frac{\sqrt[]{x^2+1}}{(x+1)^{2/3}}$
which is the first term
online calculator returned this for the answer so not sure how the 2nd term was derived the third is just 2y in numerator then simplify
$y'=\dfrac{\sqrt{x^2+1}}{\left(x+1\right)^\frac{2}{3}}
+\dfrac{x^2}{\left(x+1\right)^\frac{2}{3}\sqrt{x^2+1}}
-\dfrac{2x\sqrt{x^2+1}}{3\left(x+1\right)^\frac{5}{3}}$
I suppose you could do it that way, I would have just done...
$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= y \left[ \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3 \left( x + 1 \right) } \right] \\ &= \frac{x\,\sqrt{ x^2 + 1 }}{\left( x + 1 \right) ^{\frac{2}{3}}} \left[ \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3\left( x + 1 \right) } \right] \end{align*}$