# How do you use the chain rule to differentiate y=5tan^5(2x+1)?

##### 1 Answer
Nov 13, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 50 {\left(2 x + 1\right)}^{4} {\sec}^{2} \left(2 x + 1\right)$

#### Explanation:

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

So with $y = 5 {\tan}^{5} \left(2 x + 1\right)$, Then:

$\left\{\begin{matrix}\text{Let "u=2x+1 & => & (du)/dx=2 \\ "Let "v=tanu & => & (dv)/(du)=sec^2u \\ "Let "w=v^5 & => & (dw)/(dv)=5u^4 \\ "Then } y = 5 w & \implies & \frac{\mathrm{dy}}{\mathrm{dw}} = 5\end{matrix}\right.$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(5\right) \left(5 {u}^{4}\right) \left({\sec}^{2} u\right) \left(2\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 50 {u}^{4} {\sec}^{2} u$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 50 {\left(2 x + 1\right)}^{4} {\sec}^{2} \left(2 x + 1\right)$