Find:
$$\huge \frac d{dx} [x \tan^{-1} (e^x)]$$

Question

Find:
$$\huge \frac d{dx} [x \tan^{-1} (e^x)]$$

hartnn · Answer

didn't try product rule first
then chain rule ?

lgbasallote · Answer

product rule first then chain rule?

hartnn · Answer

can u only diff. tan^-1 (e^x) ?
or that part u need ?

hartnn · Answer

let y= tan^-1 (e^x)
d/dx(xy) = xy' + yx' = xy' + y
this product rule i was talking about....

lgbasallote · Answer

i know the product rule......but how to derive tan^-1 (e^x)

hartnn · Answer

know the derivative of tan^-1 x ?

zugzwang · Answer

Just differentiate tan^-1 x... implicitly?

lgbasallote · Answer

no idea

zugzwang · Answer

$$y = \tan^{-1}x$$
$$x = \tan y$$
Differentiate both sides with respect to x.

lgbasallote · Answer

nevermind...found it....seems using laplace transformation can work...

hartnn · Answer

d/dx (tan^-1 x) = 1/ (1+x^2)

zugzwang · Answer

And you once told me I was making things more complicated than they should be...

hartnn · Answer

O.o Laplace!

lgbasallote · Answer

that's why i made it simple @zugzwang

zugzwang · Answer

@hartnn already has the derivative...
$$x = \tan y$$$$1 = \sec^{2}y \frac{dy}{dx}$$$$\frac{dy}{dx}=\frac{1}{\sec^{2}y}$$$$\frac{dy}{dx}=\frac{1}{\tan^{2}y+1}$$
since 
x = tan y
then
$$\frac{dy}{dx}=\frac{1}{x^{2}+1}$$