# 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

$$V'(t)$$

$$V'(t)=\frac{1}{\ln (t^2+\ln (5t))}[(e^{2t}+t\cdot 2e^{2t})-\frac{te^{2t}}{t^2+\ln (5t)}(2t+\frac{1}{t})]$$

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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