How do you differentiate f(x) = x/sqrt(arctan(e^(x-1))  using the chain rule?

Aug 12, 2017

$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}\right] = \textcolor{b l u e}{\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - \frac{x {e}^{x - 1}}{2 \left(1 + {e}^{2 x - 2}\right) {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}}$

Explanation:

We're asked to find the derivative

$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}\right]$

Let's first use the product rule, which is

$\frac{d}{\mathrm{dx}} \left[u v\right] = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

where

• $u = x$

• $v = \frac{1}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}$:

$= \frac{\frac{d}{\mathrm{dx}} \left[x\right]}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right]$

The derivative of $x$ is $1$:

$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right]$

Now let's use the chain rule, which will be

$\frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right] = \frac{d}{\mathrm{du}} \left[\frac{1}{\sqrt{u}}\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = {\tan}^{-} 1 \left[{e}^{x - 1}\right]$

• $\frac{d}{\mathrm{du}} \left[\frac{1}{\sqrt{u}}\right] = - \frac{1}{2 {u}^{3 \text{/} 2}}$:

$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \left(- \frac{\frac{d}{\mathrm{dx}} \left[{\tan}^{-} 1 \left[{e}^{x - 1}\right]\right]}{2 {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}\right)$

Now we'll use the chain rule again:

$\frac{d}{\mathrm{dx}} \left[{\tan}^{-} 1 \left[{e}^{x - 1}\right]\right] = \frac{d}{\mathrm{du}} \left[{\tan}^{-} 1 u\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = {e}^{x - 1}$

• $\frac{d}{\mathrm{du}} \left[{\tan}^{-} 1 u\right] = \frac{1}{1 + {u}^{2}}$:

$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - \left(\frac{\frac{d}{\mathrm{dx}} \left[{e}^{x - 1}\right]}{{e}^{2 x - 2} + 1}\right) \frac{x}{2 {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}$

Finally, we'll use the chain rule a third time:

$\frac{d}{\mathrm{dx}} \left[{e}^{x - 1}\right] = \frac{d}{\mathrm{du}} \left[{e}^{u}\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = x - 1$

• $\frac{d}{\mathrm{du}} \left[{e}^{u}\right] = {e}^{u}$:

$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - {e}^{x - 1} \frac{d}{\mathrm{dx}} \left[x - 1\right] \frac{x}{2 \left(1 + {e}^{2 x - 2}\right) {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}$

The derivative of $x - 1$ is $1$:

color(blue)(ulbar(|stackrel(" ")(" "= 1/sqrt(tan^-1[e^(x-1)]) - (xe^(x-1))/(2(1+e^(2x-2))tan^-1[e^(x-1)]^(3"/"2)))" ")|)