Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6498

Exercise

Given the differentiable function

$z=\frac{y}{f(x^2-y^2)}$

Prove the equation

$\frac{1}{x}\cdot z'_x+\frac{1}{y}\cdot z'_y=\frac{z}{y^2}$

Proof

Define

$u=x^2-y^2$

We get the function

$z=\frac{y}{f(u)}=\frac{y}{f}$

And the internal function

$u(x,y)=x^2-y^2$

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

$z'_x=z'_f\cdot f'_u\cdot u'_x=$

$=y\cdot \frac{-1}{f^2(u)}\cdot f'_u\cdot 2x=$

$=\frac{-2xyf'_u}{f^2(u)}$

$z'_y=\frac{1\cdot f(u)-y\cdot f'_u\cdot u'_y}{f^2(u)}=$

$=\frac{1}{f(u)}+\frac{2y^2f'_u}{f^2(u)}$

We will put the partial derivatives in the left side of the equation we need to prove.

$\frac{1}{x}\cdot z'_x+\frac{1}{y}\cdot z'_y=$

$=\frac{1}{x}\cdot \frac{-2xyf'_u}{f^2(u)}+\frac{1}{y}\cdot (\frac{1}{f(u)}+\frac{2y^2f'_u}{f^2(u)})=$

$=\frac{-2yf'_u}{f^2(u)}+\frac{1}{yf(u)}+\frac{2yf'_u}{f^2(u)}=$

$=\frac{1}{yf(u)}$

Now, we will put the partial derivatives in the right side of the equation we need to prove.

$\frac{z}{y^2}=\frac{\frac{y}{f(u)}}{y^2}=$

$=\frac{y}{f(u)}\cdot\frac{1}{y^2}=\frac{1}{yf(u)}$

We got the same result in both sides as required.

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