242.7x.19 d/dx arctan(sqrt(3x))

[karush]

$\tiny{242.7x.19}$

find y'
\begin{align*}\displaystyle
y&=\tan^{-1}\sqrt{3x}
\end{align*}
$\textsf{thus would be.}$
\begin{align*}
\displaystyle
y'&=\frac{1}{(\sqrt{3x})^2 + 1} \cdot \frac{d}{dx} \sqrt{3x}
\end{align*}

$\textit{tried to follow an example but not sure about this??}$

 

[Greg]

That's correct.

 

[Prove It]

$\tiny{242.7x.19}$

find y'
\begin{align*}\displaystyle
y&=\tan^{-1}\sqrt{3x}
\end{align*}
$\textsf{thus would be.}$
\begin{align*}
\displaystyle
y'&=\frac{1}{(\sqrt{3x})^2 + 1} \cdot \frac{d}{dx} \sqrt{3x}
\end{align*}

$\textit{tried to follow an example but not sure about this??}$

I would be more inclined to use implicit differentiation in this case...

$\displaystyle \begin{align*} y &= \arctan{ \left( \sqrt{3\,x} \right) } \\ \tan{ \left( y \right) } &= \sqrt{3\,x} \\ \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \tan{(y)} \right] &= \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \left( 3\,x \right) ^{\frac{1}{2}} \right] \\ \frac{\mathrm{d}}{\mathrm{d}y}\,\left[ \tan{(y)} \right] \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 3 \cdot \frac{1}{2}\,\left( 3\,x \right) ^{-\frac{1}{2}} \\ \sec^2{(y)}\,\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2\,\sqrt{3\,x}} \\ \left[ 1 + \tan^2{(y)} \right] \,\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2\,\sqrt{3\,x}} \\ \left[ 1 + \left( \sqrt{3\,x} \right) ^2 \right] \,\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2\,\sqrt{3\,x}} \\ \left( 1 + 3\,x \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2\,\sqrt{3\,x}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2\,\sqrt{3\,x} \,\left( 1 + 3\,x \right) } \end{align*}$