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

Jun 4, 2016

$= \frac{2 {x}^{2} \cos \left(x\right) \ln \left({\cos}^{2} \left({x}^{2}\right)\right) - \ln \left({\sin}^{2} \left(x\right)\right) \setminus \sin \left(x\right) \setminus \left(2 x \ln \left({\cos}^{2} \left({x}^{2}\right)\right) - \frac{4 {x}^{3} \sin t \left({x}^{2}\right)}{\setminus \cos \left({x}^{2}\right)}\right)}{{x}^{4} {\ln}^{2} \left({\cos}^{2} \left({x}^{2}\right)\right) \sin \left(x\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \setminus \left(\setminus \frac{\setminus \ln \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right)}{{x}^{2} \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right)} \setminus\right)$

Applying quotient rule, ${\left(\setminus \frac{f}{g} \setminus\right)}^{'} = \setminus \frac{{f}^{'} \setminus \cdot g - {g}^{'} \setminus \cdot f}{{g}^{2}}$

$= \setminus \frac{\setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \ln \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right) \setminus\right) {x}^{2} \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right) - \setminus \frac{d}{\mathrm{dx}} \setminus \left({x}^{2} \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right) \setminus\right) \setminus \ln \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right)}{\setminus {\left({x}^{2} \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right) \setminus\right)}^{2}}$.............equation (i)

Now,
$\setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \ln \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right) \setminus\right)$$= \frac{2 \cos \left(x\right)}{\sin} \left(x\right)$
Applying chain rule,$\setminus \frac{\mathrm{df} \setminus \left(u \setminus\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$
$L e t {\sin}^{2} \left(x\right) = u$
$= \setminus \frac{d}{\mathrm{du}} \setminus \left(\setminus \ln \setminus \left(u \setminus\right) \setminus\right) \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right)$

and we know,
$= \setminus \frac{d}{\mathrm{du}} \setminus \left(\setminus \ln \setminus \left(u \setminus\right) \setminus\right) = \frac{1}{u}$
$= \frac{d}{\mathrm{dx}} \setminus \left(\setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus\right) = 2 \sin \left(x\right) \cos \left(x\right)$

so,$\frac{1}{u} 2 \sin \left(x\right) \cos \left(x\right)$
substituting back ${\sin}^{2} \left(x\right) = u$ we get,
$\frac{1}{\sin} ^ 2 \left(x\right) \left(2 \sin \left(x\right) \cos \left(x\right)\right)$
$= \frac{2 \cos \left(x\right)}{\sin} \left(x\right)$

Again,
\frac{d}{dx}$$x^2\ln \(\cos ^2\(x^2$$)
Applying product rule, $\setminus {\left(f \setminus \cdot g \setminus\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$
$f = {x}^{2} , \setminus g = \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right)$

=\frac{d}{dx}$$x^2$$\ln $$\cos ^2\(x^2$$\)+\frac{d}{dx}$$\ln \\cos ^2\(x^2$$\)\)x^2
We know,
$= \setminus \frac{d}{\mathrm{dx}} \setminus \left({x}^{2}\right) = 2 x$ and:
$\setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \ln \setminus \setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right) \setminus$

Applying chain rule,$\setminus \frac{\mathrm{df} \setminus \left(u \setminus\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$
$L e t {\cos}^{2} {\left(x\right)}^{2} = u$
$= \setminus \frac{d}{\mathrm{du}} \setminus \left(\setminus \ln \setminus \left(u \setminus\right) \setminus\right) \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right)$
$= \setminus \frac{1}{u} \setminus \left(- 4 x \setminus \cos \setminus \left({x}^{2} \setminus\right) \setminus \sin \setminus \left({x}^{2} \setminus\right) \setminus\right)$
Substituting back ${\cos}^{2} {\left(x\right)}^{2} = u$
$= \setminus \frac{1}{\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right)} \setminus \left(- 4 x \setminus \cos \setminus \left({x}^{2} \setminus\right) \setminus \sin \setminus \left({x}^{2} \setminus\right) \setminus\right)$
simplifying it,we get,
$\frac{- 4 x \left({\sin}^{{x}^{2}}\right)}{\cos} \left({x}^{2}\right)$

finally from equation (i),
$= 2 x \setminus \ln \setminus \left(\setminus {\cos}^{2} \setminus \left({x}^{2} \setminus\right) \setminus\right) + \setminus \left(- \setminus \frac{4 x \setminus \sin \setminus \left({x}^{2} \setminus\right)}{\setminus \cos \setminus \left({x}^{2} \setminus\right)} \setminus\right) {x}^{2}$

simplifying it,we get,
$= \frac{2 {x}^{2} \cos \left(x\right) \ln \left({\cos}^{2} \left({x}^{2}\right)\right) - \ln \left({\sin}^{2} \left(x\right)\right) \setminus \sin \left(x\right) \setminus \left(2 x \ln \left({\cos}^{2} \left({x}^{2}\right)\right) - \frac{4 {x}^{3} \sin t \left({x}^{2}\right)}{\setminus \cos \left({x}^{2}\right)}\right)}{{x}^{4} {\ln}^{2} \left({\cos}^{2} \left({x}^{2}\right)\right) \sin \left(x\right)}$