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

1 Answer
Jun 4, 2016

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)}#

Explanation:

#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)}#