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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

Final Answer

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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