# Differentiate 1/sinx^2 using chain rule?

May 10, 2017

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

#### Explanation:

We can use the chain rule to differentiate the function, the chain rule states $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

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

May 10, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = - 2 x \cot {x}^{2} {\csc}^{2} {x}^{2}$

#### Explanation:

We use chain rule here.

Using this in order to differentiate a function of a function, say $y , = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$.

In fact if we have something like $y = f \left(g \left(h \left(x\right)\right)\right)$, we can have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{df}} \times \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}}$

Here we have $f \left(x\right) = \frac{1}{g \left(x\right)}$, where $g \left(x\right) = \sin \left(h \left(x\right)\right)$ and $h \left(x\right) = {x}^{2}$.

Hence, $\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{1}{{\sin}^{2} {x}^{2}} \times \cos {x}^{2} \times 2 x = - 2 x \cot {x}^{2} {\csc}^{2} {x}^{2}$