How do you differentiate #f(x)=(ln(sinx)^2/(x^2ln(cos^2x^2)# using the chain rule?

Question

Emilia Aguilar · Accepted Answer

#=frac{2x^2cos (x)ln (cos ^2(x^2))-ln (sin ^2(x))\sin (x)$2xln (cos ^2(x^2))-frac{4x^3sint(x^2)}{\cos (x^2)})}{x^4ln ^2(cos ^2(x^2))sin (x)}# #frac{d}{dx}$\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)}#