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Multivariable Chain Rule – Exercise 3315

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

V'(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})]

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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