# \tan^{-1}(x^2 y) = x + xy^2

## Question:

{eq}\tan^{-1}(x^2 y) = x + xy^2 {/eq}

Find {eq}\frac{dy}{dx} {/eq}

## Differentiation:

Explicit Differentiation :

In explicit differentiation we need to differentiate the equation firstly as it is. Then rearrange it in order to obtain y'.

After separating y', opposite one is the answer required.

## Answer and Explanation:

We have,

{eq}\tan^{-1}(x^2 y) = x + xy^2 {/eq}

Now,

Differentiating both sides,

{eq}(\tan^{-1}(x^2 y))' = (x + xy^2)' \\ \displaystyle \frac{1}{1 + (x^2 y)^2} (x^2 y)' = 1 + (x'y^2 + x(y^2)') \\ \displaystyle \frac{1}{1 + (x^2 y)^2} (2x y + x^2 y') = 1 + (y^2 + x(2yy')) \\ \displaystyle y' \frac{x^2 - 2xy}{1 + (x^2 y)^2} = 1 + y^2 - \frac{1}{1 + (x^2 y)^2} (2x y) \\ \displaystyle y' = \frac{1 + (x^2 y)^2}{x^2 - 2xy}(1 + y^2 - \frac{2xy}{1 + (x^2 y)^2} ) \\ {/eq}

so,

{eq}\therefore \color{blue}{\displaystyle y' = \frac{1 + (x^2 y)^2}{x^2 - 2xy}(1 + y^2 - \frac{2xy}{1 + (x^2 y)^2} ) } {/eq}