 # 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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