is there any way to simplify my final expression?

Question

Given $f(x,y) = (1+y^2)\sin ^2 x $ , $ x= \tan^{-1} u $ , $y = \sin^{-1} u$

Find $df/du$

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

$\frac{\partial}{\partial y} = 2y \sin^{2} x$

$\frac{dx}{du} = \frac{1}{1+x^2} $

$\frac{dy}{du} = \frac{1}{\sqrt{1-x^2}}$

$\frac{df}{du} = \frac{\partial f}{\partial x} \cdot \frac{dx}{du} + \frac{\partial f}{\partial y} \cdot \frac{dy}{du} $

$\frac{df}{du} = 2(1+ (\sin^{-1} u)^2 ) \sin (\tan^{-1} u) \cos (\tan^{-1} u) \cdot \frac{1}{1+ (\tan^{-1} u)^2} + (2(\sin^{-1} \sin^2 (\tan^{-1} u) \cdot \frac{1}{\sqrt{1- (\tan^{-1})^2}} $

Just thought that this is a weird expression and thus cannot be simplified. Am I right ?

Answer 1

One thing you could do is you know that $$1+\frac{1}{\tan^2{\alpha}}=\frac{1}{\sin^2{\alpha}}$$ Put $\alpha = \tan^{-1}{u}$ and youll get: $$\sin^{2}{(\tan^{-1}{u})}=\frac{u^2}{1+u^2}$$

Take the square root of this, and you'll have another expression of yours.

All of this is derived playing with the equation: $$\sin^2{\alpha}+\cos^2{\alpha}=1$$

Answer 2

Alternatively: $$f(u)=(1+\arcsin^2 u)\cdot \sin^2 (\arctan u)\\ f'(u)=2\arcsin u\cdot \frac{1}{\sqrt{1-u^2}}\cdot \sin^2 (\arctan u)+(1+\arcsin^2u)\cdot \sin(2\arctan u)\cdot \frac{1}{1+u^2}=\\ 2\arcsin u\cdot \frac{1}{\sqrt{1-u^2}}\cdot \frac{u^2}{1+u^2}+(1+\arcsin^2u)\cdot \frac{2u}{1+u^2}\cdot \frac{1}{1+u^2}=\\ \frac{2u^2\arcsin u}{\sqrt{1-u^2}\cdot (1+u^2)}+\frac{2u(1+\arcsin^2u)}{(1+u^2)^2}.$$ See WA result.

Note: $$\sin^2(\arctan u)=1-\cos^2(\arctan u)=1-\frac{1}{1+\tan^2(\arctan u)}=1-\frac{1}{1+u^2}=\frac{u^2}{1+u^2};\\ \sin(\arctan u)=\frac{u}{\sqrt{1+u^2}};\\ \cos^2(\arctan u)=\frac{1}{1+\tan^2(\arctan u)}=\frac{1}{1+u^2};\\ \cos(\arctan u)=\frac{1}{\sqrt{1+u^2}};\\ \sin(2\arctan u)=2\sin(\arctan u)\cos (\arctan u)=\frac{2u}{1+u^2}.$$