# How do you differentiate f(x)= ln(sin(x^2)/x) ?

Dec 30, 2015

To differentiate the $\ln$, we'll need quotient rule.
To differentiate $\sin \left({x}^{2}\right)$, we'll need chain rule as well.
To differentiate $\sin \frac{{x}^{2}}{x}$, we'll need quotient rule.

#### Explanation:

• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
• Quotient rule: be $y = f \frac{x}{g} \left(x\right)$, then $y ' = \frac{f ' g - f g '}{g} ^ 2$

We can rename $u = \sin \frac{{x}^{2}}{x}$ so that $f \left(x\right) = \ln \left(u\right)$, which is differentiable.

Also, we can rename $v = {x}^{2}$ so we can differentiate $\sin \left(v\right)$ applying chain rule, as well.

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

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

Recalling trigonometric identities: $\cot a = \cos \frac{a}{\sin} a$

Let's split the result a bit.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{\cancel{2}} \textcolor{g r e e n}{\cos \left({x}^{2}\right)}}{\cancel{x} \textcolor{g r e e n}{\sin \left({x}^{2}\right)}} - \frac{\cancel{\left(\sin \left({x}^{2}\right)\right)}}{x \cancel{\sin \left({x}^{2}\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \cot \left({x}^{2}\right) - \frac{1}{x}$