MHB 242.ws3.1 d/dx of e^arctan{x} - arctan{e^x}

[karush]

$\large{242.ws3.1}\\$
$\d{}{x}e^{x}=e^{x}$ and $\d{}{x}\left[\tan^{-1}\left({x}\right)\right]=\frac{1}{1+{x}^{2}}$

find the derivative
$$\d{}{x} \left[e^{\arctan{x}} - \arctan{(e^x)}\right]
=\dfrac{\mathrm{e}^{\arctan\left(x\right)}}{x^2+1}-\dfrac{\mathrm{e}^x}{\mathrm{e}^{2x}+1}$$

tried but got lost

 

[MarkFL]

I think what I would do is begin by defining:

$$f=e^{\arctan(x)}$$

Convert from exponential to logarithmic form:

$$\ln(f)=\arctan(x)$$

Now, implicitly differentiate:

$$\frac{f'}{f}=\frac{1}{x^2+1}$$

Hence:

$$f'=\frac{f}{x^2+1}=\frac{e^{\arctan(x)}}{x^2+1}$$

Okay, next let's define:

$$g=\arctan\left(e^x\right)$$

Take the tangent of both sides:

$$\tan(g)=e^x$$

Implicitly differentiate:

$$\sec^2(g)g'=e^x$$

Hence:

$$g'=\frac{e^x}{\sec^2(g)}=\frac{e^x}{\tan^2(g)+1}=\frac{e^x}{e^{2x}+1}$$

And so we have:

$$\frac{d}{dx}(f-g)=f'-g'=\frac{e^{\arctan(x)}}{x^2+1}-\frac{e^x}{e^{2x}+1}$$