B Really an easy question about derivative of arctan

[kidsasd987]

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/MIT18_01SCF10_Ses15b.pdfso derivative of arctan is 1/(x^2+1) and this is obvious since we are considering trigonometry of a triangle of which hypotenuse is sqrt(x^2+1).But can we consider its hypotenuse to be x instead of sqrt(x^2+1)? where cosy=1/x

 

[member 587159]

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/MIT18_01SCF10_Ses15b.pdfso derivative of arctan is 1/sqrt(x^2+1) and this is obvious since we are considering trigonometry of a triangle of which hypotenuse is sqrt(x^2+1).But can we consider its hypotenuse to be x instead of sqrt(x^2+1)? where cosy=1/x

$\frac{d \arctan(x)}{dx} = \frac{1}{1+x^2}$, and not what you say.

 

[kidsasd987]

$\frac{d \arctan(x)}{dx} = \frac{1}{1+x^2}$, and not what you say.

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 1Ah please don't mind. I figured out.

 

[Mark44]

But can we consider its hypotenuse to be x instead of sqrt(x^2+1)? where cosy=1/x

Ah please don't mind. I figured out.

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 $\cos^2(y) = \frac 1 {x^2 + 1}$