Exercise
Given
v(x,y)=\frac{x}{y}
f(t)=te^{2t}
g(t)=\ln (t^2+\ln (5t))
V(t)=v(f(t),g(t))
Calculate the derivative
V'(t)
Final Answer
Solution
We will use the chain rule to calculate the derivative
V'(t)=v'_f\cdot f'_t+v'_g\cdot g'_t
We have
V(t)=v(f(t),g(t))
And function v is
v(x,y)=\frac{x}{y}
We put it in function u and get
u(f,g)=\frac{f}{g}
We calculate the patial derivatives of v.
v'_f=\frac{1}{g}
v'_g=\frac{-f}{g^2}
Also, we calculate the derivatives of f and g.
f'_t=e^{2t}+t\cdot 2e^{2t}
g'_t=\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{5t}\cdot 5)=
=\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t})
We put the derivatives in V’ and get
V'(t)=v'_f\cdot f'_t+v'_g\cdot g'_t=
=\frac{1}{g}\cdot(e^{2t}+t\cdot 2e^{2t})+\frac{-f}{g^2}\cdot\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t})=
=\frac{1}{\ln (t^2+\ln (5t))}\cdot(e^{2t}+t\cdot 2e^{2t})+\frac{-te^{2t}}{\ln^2 (t^2+\ln (5t))}\cdot\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t})=
=\frac{1}{\ln (t^2+\ln (5t))}[(e^{2t}+t\cdot 2e^{2t})-\frac{te^{2t}}{t^2+\ln (5t)}(2t+\frac{1}{t})]
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