How do you differentiate f(x)=sinx^2/x^2?

May 22, 2017

Use the quotient rule.

Explanation:

For $f \left(x\right) = \frac{u}{v}$, we have $f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$

In this quotient, $u = \sin {x}^{2}$, which I will take to be $\sin \left({x}^{2}\right)$ (rather than ${\left(\sin x\right)}^{2}$ which is a different function).

This makes $u ' = \cos \left({x}^{2}\right) \cdot 2 x = 2 x \cos \left({x}^{2}\right)$ $\text{ }$ (chain rule)

and $v = {x}^{2}$, so $v ' = 2 x$.

$f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$

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

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

Every term has a factor of $x$, so we can simplify the answer.

$= \frac{\cancel{x} \left(2 {x}^{2} \cos \left({x}^{2}\right) - 2 \sin \left({x}^{2}\right)\right)}{\cancel{x} {x}^{3}}$

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