Vtyjkyuk

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

Find $df/du$

$fracpartialpartial x = 2(1+y^2) sin x cos x $

$fracpartialpartial y = 2y sin^2 x$

$fracdxdu = frac11+x^2 $

$fracdydu = frac1sqrt1-x^2$

$fracdfdu = fracpartial fpartial x cdot fracdxdu + fracpartial fpartial y cdot fracdydu $

$fracdfdu = 2(1+ (sin^-1 u)^2 ) sin (tan^-1 u) cos (tan^-1 u) cdot frac11+ (tan^-1 u)^2 + (2(sin^-1 sin^2 (tan^-1 u) cdot frac1sqrt1- (tan^-1)^2 $

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

One thing you could do is you know that $$1+frac1tan^2alpha=frac1sin^2alpha$$
Put $alpha = tan^-1u$ and youll get:
$$sin^2(tan^-1u)=fracu^21+u^2$$

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

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

Alternatively:
$$f(u)=(1+arcsin^2 u)cdot sin^2 (arctan u)\
f'(u)=2arcsin ucdot frac1sqrt1-u^2cdot sin^2 (arctan u)+(1+arcsin^2u)cdot sin(2arctan u)cdot frac11+u^2=\
2arcsin ucdot frac1sqrt1-u^2cdot fracu^21+u^2+(1+arcsin^2u)cdot frac2u1+u^2cdot frac11+u^2=\
frac2u^2arcsin usqrt1-u^2cdot (1+u^2)+frac2u(1+arcsin^2u)(1+u^2)^2.$$
See WA result.

Note:
$$sin^2(arctan u)=1-cos^2(arctan u)=1-frac11+tan^2(arctan u)=1-frac11+u^2=fracu^21+u^2;\
sin(arctan u)=fracusqrt1+u^2;\
cos^2(arctan u)=frac11+tan^2(arctan u)=frac11+u^2;\
cos(arctan u)=frac1sqrt1+u^2;\
sin(2arctan u)=2sin(arctan u)cos (arctan u)=frac2u1+u^2.$$