# How do you differentiate y=tan^-1(3x)?

##### 1 Answer
Aug 12, 2016

$= \frac{3}{9 {x}^{2} + 1}$

#### Explanation:

for $\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(3 x\right)\right)$

you can remember that

$\frac{d}{\mathrm{du}} \left({\tan}^{- 1} u\right) = \frac{1}{1 + {u}^{2}}$

and that, where $u = u \left(v\right)$, via the chain rule:

$\frac{d}{\mathrm{dv}} \left({\tan}^{- 1} u\right) = \frac{1}{1 + {u}^{2} \left(u\right)} \cdot \frac{\mathrm{du}}{\mathrm{dv}}$

or you can switch the function over by saying that

$\tan y = 3 x$ and then differentiating implicitly, so that

${\sec}^{2} y \setminus y ' = 3$

BTW you are still using the chain rule because:

$\frac{d \left(\tan y\right)}{\mathrm{dx}} = \frac{d \left(\tan y\right)}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

All of which means that:

$\setminus y ' = \frac{3}{\sec} ^ 2 y$

$\setminus y ' = \frac{3}{{\tan}^{2} y + 1}$

$\setminus y ' = \frac{3}{{\left(3 x\right)}^{2} + 1}$

$= \frac{3}{9 {x}^{2} + 1}$