# How do you find the derivative of arctan(x^2y)?

##### 1 Answer
May 28, 2018

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

#### Explanation:

So, basically, you want to find $\frac{d}{\mathrm{dx}} \left(\arctan \left({x}^{2} y\right)\right)$.

We need to first observe that $y$ and $x$ have no relation to each other in the expression. This observation is very important, since now $y$ can be treated as a constant with respect to $x$.

We first apply chain rule:
$\frac{d}{\mathrm{dx}} \left(\arctan \left({x}^{2} y\right)\right) = \frac{d}{d \left({x}^{2} y\right)} \left(\arctan \left({x}^{2} y\right)\right) \times \frac{d}{\mathrm{dx}} \left({x}^{2} y\right) = \frac{1}{1 + {\left({x}^{2} y\right)}^{2}} \times \frac{d}{\mathrm{dx}} \left({x}^{2} y\right)$.

Here, as we mentioned earlier, $y$ is a constant with respect to $x$. So,

$\frac{d}{\mathrm{dx}} \left({x}^{2} \textcolor{red}{y}\right) = \textcolor{red}{y} \times \frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x y$

So, $\frac{d}{\mathrm{dx}} \left(\arctan \left({x}^{2} y\right)\right) = \frac{1}{1 + {\left({x}^{2} y\right)}^{2}} \times 2 x y = \frac{2 x y}{1 + {\left({x}^{2} y\right)}^{2}}$