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
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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