#frac{d}{dx}(\frac{\ln (\sin ^2(x))}{x^2\ln (\cos ^2(x^2))})#
Applying quotient rule, #(\frac{f}{g})^'=\frac{f^'\cdot g-g^'\cdot f}{g^2}#
#=\frac{\frac{d}{dx}(\ln (\sin ^2(x)))x^2\ln (\cos ^2(x^2))-\frac{d}{dx}(x^2\ln (\cos ^2(x^2)))\ln (\sin ^2(x))}{(x^2\ln (\cos ^2(x^2)))^2}#.............equation (i)
Now,
#\frac{d}{dx}(\ln (\sin ^2(x)))##=(2cos(x))/sin(x)#
Applying chain rule,#\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}#
#Let sin^2 (x) = u#
#=\frac{d}{du}(\ln (u))\frac{d}{dx}(\sin ^2(x))#
and we know,
#=\frac{d}{du}(\ln (u)) = 1/u#
#=frac{d}{dx}(\sin ^2(x)) =2sin(x)cos(x)#
so,#1/u 2sin(x)cos(x)#
substituting back #sin^2 (x) = u# we get,
#1/sin^2(x) (2sin(x) cos(x))#
#=(2cos(x))/sin(x)#
Again,
#\frac{d}{dx}(x^2\ln (\cos ^2(x^2))#
Applying product rule, #(f\cdot g)^'=f^'\cdot g+f\cdot g^'#
#f=x^2,\g=\ln (\cos ^2(x^2))#
#=\frac{d}{dx}(x^2)\ln (\cos ^2(x^2))+\frac{d}{dx}(\ln \cos ^2(x^2)))x^2#
We know,
#=\frac{d}{dx}(x^2)=2x# and:
#\frac{d}{dx}(\ln \cos ^2(x^2))#
Applying chain rule,#\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}#
#Let cos^2(x)^2 = u#
#=\frac{d}{du}(\ln (u))\frac{d}{dx}(\cos ^2(x^2))#
#=\frac{1}{u}(-4x\cos (x^2)\sin (x^2))#
Substituting back #cos^2(x)^2 = u#
#=\frac{1}{\cos ^2(x^2)}(-4x\cos (x^2)\sin (x^2))#
simplifying it,we get,
#(-4x(sin^(x^2)))/cos(x^2)#
finally from equation (i),
#=2x\ln (\cos ^2(x^2))+(-\frac{4x\sin (x^2)}{\cos (x^2)})x^2#
simplifying it,we get,
#=frac{2x^2cos (x)ln (cos ^2(x^2))-ln (sin ^2(x))\sin (x)(2xln (cos ^2(x^2))-frac{4x^3sint(x^2)}{\cos (x^2)})}{x^4ln ^2(cos ^2(x^2))sin (x)}#
