Question

Can someone explain how to find: d\dx tan^-1.

Answer 1

Let's say the inverse tan of x gives us an angle theta:
\[ \tan^{-1}x = \theta \]
take the derivative with respect to "x", of both sides:
\[ \frac{d}{dx}\left( \tan^{-1}x\right) = \frac{d \theta}{dx} \]
this tells us we want to find d\(\theta\)/dx.

How do do that ?
if we start with the original problem, and take the tangent of both sides, like so:
\[ \tan^{-1}x = \theta \\ x = \tan\theta\]
now take the derivative of both sides with respect to theta (you are able to find the derivative of tan, right?)
\[ \frac{dx}{d\theta} = \sec^2 \theta = \frac{1}{\cos^2 \theta} \]
and if we "flip" both sides:
\[ \frac{d\theta}{dx}= \cos^2 \theta \]
now we have d\(\theta\)/dx, so we can say
\[ \frac{d}{dx}\left( \tan^{-1}x\right) =\frac{d\theta}{dx}= \cos^2 \theta \]

that is the answer, except we want the derivative in terms of x and not \(\theta\)

Answer 2

to replace the theta, we draw a triangle with angle theta, and sides such that the tangent of theta is x. A reasonable choice of lengths would be:


we see that
\[ \cos \theta = \frac{1}{\sqrt{x^2+1}} \\ \]
and
\[ \cos^2 \theta = \frac{1}{x^2+1} \]
and using this result, we get the final answer:
\[ \frac{d}{dx} \tan^{-1} x = \frac{1}{x^2+1} \]