Q. 28 - Calculus

Question

Find the first-order partial derivatives for the functions in Exercises 27–36. 

$g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right)$

Step-by-Step Solution

Answer

The first-order partial derivatives are 

$\frac{\partial g}{\partial x}=\frac{y^2}{1+x^2{y}^4} \text{and} \frac{\partial g}{\partial y}=\frac{2xy}{1+x^2{y}^4}$

Step 1: Given

The given function is: $g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right)$.

Step 2: To find

We have to find the first-order partial derivatives. 

Step 3: Calculation

$$\begin{gathered} g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right) \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left({\tan }^{-1}\left(x{y}^2\right)\right) \\ \frac{\partial g}{\partial x}=\frac{1}{1+{\left(x{y}^2\right)}^2}\frac{\partial }{\partial x}\left(x{y}^2\right) \\ \frac{\partial g}{\partial x}=\frac{1}{1+x^2{y}^4}\left(y^2\right) \\ \frac{\partial g}{\partial x}=\frac{y^2}{1+x^2{y}^4} \\ g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right) \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left({\tan }^{-1}\left(x{y}^2\right)\right) \\ \frac{\partial g}{\partial y}=\frac{1}{1+{\left(x{y}^2\right)}^2}\frac{\partial }{\partial y}\left(x{y}^2\right) \\ \frac{\partial g}{\partial y}=\frac{1}{1+x^2{y}^4}\left(2xy\right) \\ \frac{\partial g}{\partial y}=\frac{2xy}{1+x^2{y}^4} \\ \text{Hence,} \frac{\partial g}{\partial x}=\frac{y^2}{1+x^2{y}^4} \text{and} \frac{\partial g}{\partial y}=\frac{2xy}{1+x^2{y}^4} \end{gathered}$$