Q. 29 - Calculus

Question

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

$f\left(x,y\right)=x^y$

Step-by-Step Solution

Answer

The first-order partial derivatives are  $\frac{\partial f}{\partial x}=y{x}^{\left(y-1\right)} \text{and }\frac{\partial f}{\partial y}=x^y\ln \left(x\right)$

Step 1: Given

The given function is:  $f\left(x,y\right)=x^y$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

$$\begin{gathered} f\left(x,y\right)=x^y \\ \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(x^y\right) \\ \frac{\partial f}{\partial x}=y{x}^{\left(y-1\right)} \\ f\left(x,y\right)=x^y \\ \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(x^y\right) \\ \frac{\partial f}{\partial y}=x^y\ln \left(x\right) \\ \text{Hence,} \frac{\partial f}{\partial x}=y{x}^{\left(y-1\right)} \text{and }\frac{\partial f}{\partial y}=x^y\ln \left(x\right) \end{gathered}$$