Multivariable Chain Rule – Exercise 3324

Exercise

Given

$$z(u,v)=u^2\ln v$$

$$f(x,y)=\frac{x}{y}$$

$$g(x,y)=\frac{y}{x}$$

$$Z(x,y)=z(f(x,y),g(x,y))$$

Calculate the derivative

$$Z'_x,Z'_y$$

$$Z'_x=\frac{2x}{y^2}\ln \frac{y}{x}-\frac{x}{y^2}$$

$$Z'_y=\frac{-2x^2}{y^3}\ln \frac{y}{x}+\frac{x^2}{y^3}$$

Solution

We will use the chain rule to calculate the derivative

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

We have

$$Z(x,y)=z(f(x,y),g(x,y))$$

And function z is

$$z(u,v)=u^2\ln v$$

Hence, we get

$$z(f,g)=f^2\ln g$$

We calculate the partial derivatives of z.

$$z'_f=2f\ln g$$

$$z'_g=\frac{f^2}{g}$$

And also the derivatives of f and g.

$$f'_x=\frac{1}{y}$$

$$g'_x=\frac{-y}{x^2}$$

We put the derivatives in Z’ and get

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

$$=2f\ln g\cdot \frac{1}{y}+\frac{f^2}{g}\cdot\frac{-y}{x^2}=$$

$$=2\frac{x}{y}\ln \frac{y}{x}\cdot \frac{1}{y}+\frac{{(\frac{x}{y})}^2}{\frac{y}{x}}\cdot\frac{-y}{x^2}=$$

$$=\frac{2x}{y^2}\ln \frac{y}{x}-\frac{x^3}{y^3}\cdot\frac{y}{x^2}=$$

$$=\frac{2x}{y^2}\ln \frac{y}{x}-\frac{x}{y^2}$$

We will use the chain rule to calculate the derivative

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

Also we got above the derivatives

$$z'_f=2f\ln g$$

$$z'_g=\frac{f^2}{g}$$

We calculate the derivatives of f and g

$$f'_y=\frac{-x}{y^2}$$

$$g'_y=\frac{1}{x}$$

We put the derivatives in Z’ and get

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

$$=2f\ln g\cdot \frac{-x}{y^2}+\frac{f^2}{g}\cdot\frac{1}{x}=$$

$$=2\frac{x}{y}\ln \frac{y}{x}\cdot \frac{-x}{y^2}+\frac{{(\frac{x}{y})}^2}{\frac{y}{x}}\cdot\frac{1}{x}=$$

$$=\frac{-2x^2}{y^3}\ln \frac{y}{x}+\frac{x^3}{y^3}\cdot\frac{1}{x}=$$

$$=\frac{-2x^2}{y^3}\ln \frac{y}{x}+\frac{x^2}{y^3}$$

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