# B Really an easy question about derivative of arctan

1. Apr 13, 2017

### kidsasd987

Last edited: Apr 13, 2017
2. Apr 13, 2017

### Math_QED

3. Apr 13, 2017

### kidsasd987

whoops sorry somehow I put it in sqrt.

Can you explain me why we are considering one specific trigonometry although we could consider a triangle of which hypotenuse is x and base is 1

Ah please don't mind. I figured out.

Last edited: Apr 13, 2017
4. Apr 13, 2017

### Staff: Mentor

I'm not sure what it is that you figured out, but here's the answer to your first question, above.
Let $y = \arctan(x)$
An equivalent equation is $\tan(y) = x$
The picture below shows a right triangle with one angle labeled y. Since $\tan(y) = x$, it's reasonable to label the opposite side as x, and the adjacent side as 1. This forces the hypotenuse to be $\sqrt{x^2 + 1}$.

If we differentiate both sides of the equation $\tan(y) = x$ with respect to x, we get $\sec^2(y) \frac{dy}{dx} = 1$, or $\frac{dy}{dx} = \frac 1{\sec^2(y)} = \cos^2(y)$. From the drawing, it can be seen that $\cos^2(y) = \frac 1 {x^2 + 1}$