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

1 Answer
Jun 4, 2016

=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)}