# How do you differentiate f(x)=e^cot(sqrt(x))  using the chain rule?

Mar 23, 2017

$f ' \left(x\right) = - \frac{{\csc}^{2} \left(\sqrt{x}\right) \cdot {e}^{\cot \sqrt{x}}}{2 \sqrt{x}}$

#### 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 $y = f \left(x\right) = {e}^{\cot \sqrt{x}}$, Then:

 { ("Let",u=sqrt(x)=x^(1/2), => , (du)/dx=1/2x^(-1/2)=1/(2sqrt(x))), ("And",v=cot sqrt(x)=cotu, => , (dv)/(du)=-csc^2u), ("Then",y=e^(cot sqrt(x))=e^v, =>, dy/(dv)=e^v ) :}

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

$\setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left({e}^{v}\right) \left(- {\csc}^{2} u\right) \left(\frac{1}{\sqrt{x}}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({e}^{\cot \sqrt{x}}\right) \left(- {\csc}^{2} \sqrt{x}\right) \left(\frac{1}{2 \sqrt{x}}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{{\csc}^{2} \left(\sqrt{x}\right) \cdot {e}^{\cot \sqrt{x}}}{2 \sqrt{x}}$