Multivariable Chain Rule – Exercise 3350



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

f(u,v)=u\sin v

g(u,v)=u\cos v


Calculate the derivative


Final Answer




We will use the chain rule to calculate the derivative

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

We have


And function u is


We put it in the function above and get


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

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