y=sin^2(lnx^2) What is dy/dx ?

$y = {\sin}^{2} \left(\ln {x}^{2}\right)$ . Find $\frac{\mathrm{dy}}{\mathrm{dx}}$

Apr 20, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \sin \left(\ln {x}^{2}\right) \cos \left(\ln {x}^{2}\right)}{x}$

Explanation:

If $y = {\sin}^{2} \left(\ln {x}^{2}\right)$ then

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dl} n {x}^{2}} {\sin}^{2} \left(\ln {x}^{2}\right) \frac{d}{\mathrm{dx}} \ln {x}^{2} = 2 \sin \left(\ln {x}^{2}\right) \cos \left(\ln {x}^{2}\right) \frac{2 x}{x} ^ 2 = \frac{4 \sin \left(\ln {x}^{2}\right) \cos \left(\ln {x}^{2}\right)}{x}$

Apr 21, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \sin \left(\ln \left({x}^{2}\right)\right) \cos \left(\ln \left({x}^{2}\right)\right)}{x}$

Explanation:

We use the chain rule twice here:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{\sin}^{2} \left(\ln \left({x}^{2}\right)\right)\right]$

Some rules to recall here:

$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

$\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left[\sin x\right] = \cos x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sin \left(\ln \left({x}^{2}\right)\right) \cdot \frac{d}{\mathrm{dx}} \left[\sin \left(\ln \left({x}^{2}\right)\right)\right]$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sin \left(\ln \left({x}^{2}\right)\right) \cdot \cos \left(\ln \left({x}^{2}\right)\right) \cdot \frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2}\right)\right]$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sin \left(\ln \left({x}^{2}\right)\right) \cdot \cos \left(\ln \left({x}^{2}\right)\right) \cdot \frac{1}{{x}^{2}} \cdot \frac{d}{\mathrm{dx}} \left[{x}^{2}\right]$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sin \left(\ln \left({x}^{2}\right)\right) \cdot \cos \left(\ln \left({x}^{2}\right)\right) \cdot \frac{1}{{x}^{2}} \cdot 2 x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cdot \sin \left(\ln \left({x}^{2}\right)\right) \cdot \cos \left(\ln \left({x}^{2}\right)\right) \cdot \frac{1}{x}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \sin \left(\ln \left({x}^{2}\right)\right) \cos \left(\ln \left({x}^{2}\right)\right)}{x}$