# Multivariable Chain Rule – Exercise 3350

Exercise

Given

$$z(x,y)=\arctan (\frac{x}{y})$$

$$f(u,v)=u\sin v$$

$$g(u,v)=u\cos v$$

$$Z(u,v)=z(f(u,v),g(u,v))$$

Calculate the derivative

$$Z'_u,Z'_v$$

$$Z'_u=0$$

$$Z'_v=1$$

Solution

We will use the chain rule to calculate the derivative

$$Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u$$

We have

$$Z(u,v)=z(f(u,v),g(u,v))$$

And function u is

$$z(x,y)=\arctan(\frac{x}{y})$$

We put it in the function above and get

$$z(f,g)=\arctan(\frac{f}{g})$$

Hence, the partial derivatives of z are

$$z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}$$

$$z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}$$

Also, we calculate the derivatives of f and g

$$f'_u=\sin v$$

$$g'_u=\cos v$$

We put all the derivatives and get

$$Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u=$$

$$=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot \sin v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot \cos v=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot \sin v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot \cos v=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}+\frac{-u\sin v}{u^2\cos^2 v}\cdot \cos v)=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}-\frac{\sin v}{u\cos v})=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot 0=0$$

We will use the chain rule to calculate the partial derivatives of Z.

$$Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v$$

We got above

$$z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}$$

$$z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}$$

Alao, we calculate the derivatives of f and g.

$$f'_v=u\cos v$$

$$g'_v=-u\sin v$$

We put the derivatives and get

$$Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v$$

$$=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot u\cos v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot(-u\sin v)=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot u\cos v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v)=$$

$$=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{u\cos v}{u\cos v} +\frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v))=$$

$$=\frac{1}{1+\frac{\sin^2 v}{\cos^2 v}}(1+\frac{u^2\sin^2 v}{u^2\cos^2 v})=$$

$$=\frac{1}{1+\tan^2 v}(1+\tan^2 v)=$$

$$=\frac{1+\tan^2 v}{1+\tan^2 v}=1$$

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

Share with Friends