# How do you find the derivative of the function y= tan^-1 (3x)?

Apr 14, 2015

Remembering that the derivative of:

$y = {\tan}^{-} 1 x$

is

$y ' = \frac{1}{1 + {x}^{2}}$

than:

$y ' = \frac{1}{1 + {\left(3 x\right)}^{2}} \cdot 3 = \frac{3}{1 + 9 {x}^{2}}$.

Apr 14, 2015

By the Chain Rule (and knowledge of the derivative of inverse tangent function), the answer is: $\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{1}{1 + {\left(3 x\right)}^{2}} \setminus \cdot 3 = \setminus \frac{3}{1 + 9 {x}^{2}}$.

If you didn't happen to remember the derivative of the inverse tangent function, you could derive the answer by differentiation of both sides of the equation $\tan \left({\tan}^{- 1} \left(3 x\right)\right) = 3 x$ with respect to $x$ to get, by the Chain Rule, ${\sec}^{2} \left({\tan}^{- 1} \left(3 x\right)\right) \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(3 x\right)\right) = 3$ so that $\setminus \frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(3 x\right)\right) = 3 {\cos}^{2} \left({\tan}^{- 1} \left(3 x\right)\right)$.

To simplify this, at this point, you could draw a right triangle and label one of the non-right angles $\setminus {\tan}^{- 1} \left(3 x\right)$. Since the tangent of the angle is the length of the opposite side divided by the length of the adjacent side, you could label the opposite side with "$3 x$" and the adjacent side with "1". The Pythagorean Theorem would imply that the length of the hypotenuse is $\setminus \sqrt{1 + 9 {x}^{2}}$. Since the cosine of the angle is the length of the adjacent side divided by the length of the hypotenuse, you'd get \cos(\tan^{-1}(3x))=\frac{1}{\sqrt{1+9x^{2}}.

Hence, $\setminus \frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(3 x\right)\right) = 3 {\cos}^{2} \left({\tan}^{- 1} \left(3 x\right)\right) = \setminus \frac{3}{1 + 9 {x}^{2}}$