# Differentiate Tan^-1y=x^3?

Sep 2, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} {\sec}^{2} \left({x}^{3}\right)$

#### Explanation:

${\tan}^{-} 1 y = {x}^{3}$

Method 1 - No simplification

Differentiate as written. Recall that $\frac{d}{\mathrm{dx}} {\tan}^{-} 1 x = \frac{1}{1 + {x}^{2}}$. Here, differentiating ${\tan}^{-} 1 y$ with respect to $x$ will cause the chain rule to be in effect.

$\frac{1}{1 + {y}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2}$

Solving for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \left(1 + {y}^{2}\right)$

Let's write this all in terms of $x$. To do this, we have to solve for $y$. From ${\tan}^{-} 1 y = {x}^{3}$, note that $y = \tan \left({x}^{3}\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \left(1 + {\tan}^{2} \left({x}^{3}\right)\right)$

Note that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} {\sec}^{2} \left({x}^{3}\right)$

Method 2 - Simplification

Recall that $y = \tan \left({x}^{3}\right)$. We can then differentiate this directly, which is easier. It helps to know that $\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$. We will use that here and also use the chain rule.

$y = \tan \left({x}^{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left({x}^{3}\right) \cdot \frac{d}{\mathrm{dx}} {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} {\sec}^{2} \left({x}^{3}\right)$