frac{d}{dx}(\frac{\ln (\sin ^2(x))}{x^2\ln (\cos ^2(x^2))})ddx(ln(sin2(x))x2ln(cos2(x2)))
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)}
