Prove the derivative of $\arctan(x)$ using derivative definition

Question

Prove that the derivative of $\arctan(x)$ is $\frac1{1+x^2}$ using definition of derivative

I'm not allowed to use derivative of inverse function, infinite series or l'Hopital.

Answer 1

You need some trigonometry. Since $\arctan (x+h)-\arctan x=\arctan\tfrac{h}{1+x(x+h)}$, $\lim_{h\to 0}\tfrac{\arctan (x+h)-\arctan x}{h}=\tfrac{1}{\lim_{h\to 0} (1+x(x+h))}$ (because $\arctan\theta\approx\theta$ for small $\theta$).