# How do you differentiate f(x)=csc(1/x^3)  using the chain rule?

##### 1 Answer
Mar 23, 2017

$f ' \left(x\right) = \frac{3 \csc \left(\frac{1}{x} ^ 3\right) \cot \left(\frac{1}{x} ^ 3\right)}{x} ^ 4$

#### Explanation:

f you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If $y = f \left(x\right)$ then $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

I was taught to remember that the differential can be treated like a fraction and that the "$\mathrm{dx}$'s" of a common variable will "cancel" (It is important to realise that $\frac{\mathrm{dy}}{\mathrm{dx}}$ isn't a fraction but an operator that acts on a function, there is no such thing as "$\mathrm{dx}$" or "$\mathrm{dy}$" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ etc, or $\left(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\textcolor{red}{\cancel{\mathrm{dv}}}} \frac{\textcolor{red}{\cancel{\mathrm{dv}}}}{\textcolor{b l u e}{\cancel{\mathrm{du}}}} \frac{\textcolor{b l u e}{\cancel{\mathrm{du}}}}{\mathrm{dx}}\right)$

So with $y = \csc \left(\frac{1}{x} ^ 3\right)$, Then:

$\left\{\begin{matrix}\text{Let "u=1/x^3 & => & (du)/dx=-3/x^4 \\ "Then } y = \csc u & \implies & \frac{\mathrm{dy}}{\mathrm{du}} = - \csc u \cot u\end{matrix}\right.$

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$ we get:

$\setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \csc u \cot u\right) \left(- \frac{3}{x} ^ 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 \csc u \cot u}{x} ^ 4$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 \csc \left(\frac{1}{x} ^ 3\right) \cot \left(\frac{1}{x} ^ 3\right)}{x} ^ 4$