Q. 40 - Calculus

Question

Find the gradient of the given functions in Exercises 37–42. 

$f\left(x,y\right)=yse{c}^{-1}\left(\frac{1}{x}\right)$

Step-by-Step Solution

Answer

The gradient of the given function is $\left(-\frac{y}{\sqrt{1-x^2}},se{c}^{-1}\left(\frac{1}{x}\right)\right).$

Step 1. Given Information.

The given function is:

$f\left(x,y\right)=yse{c}^{-1}\left(\frac{1}{x}\right)$

Step 2. Calculation.

The gradient of the given function is: 

$$\begin{gathered} z=f\left(x,y\right)=yse{c}^{-1}\left(\frac{1}{x}\right) \\ \nabla f(x,y)=\left[f_x\left(x,y\right),f_y\left(x,y\right)\right]-------\left(1\right) \end{gathered}$$

Now find  

$$\begin{gathered} f_x=\frac{\partial f(x,y)}{\partial x}=y\left(\frac{1·\left(-{\displaystyle \frac{1}{x^2}}\right)}{\left({\displaystyle \frac{1}{x}}\right)\sqrt{{\left({\displaystyle \frac{1}{x}}\right)}^2}-1}\right)=-\frac{y}{\sqrt{1-x^2}} \\ {f}_y=\frac{\partial f(x,y)}{\partial y}=se{c}^{-1}\left(\frac{1}{x}\right)·1=se{c}^{-1}\left(\frac{1}{x}\right) \end{gathered}$$

Use these above values in (1) we get 

$\nabla f(x,y)=\left(-\frac{y}{\sqrt{1-x^2}},se{c}^{-1}\left(\frac{1}{x}\right)\right)$

Step 3. Conclusion.

The gradient of the given function is $\left(-\frac{y}{\sqrt{1-x^2}},se{c}^{-1}\left(\frac{1}{x}\right)\right)$.