# Multivariable Chain Rule – Exercise 3367

Exercise

Given

$$u(x,y)=e^{xy^2}$$

$$f(t)=t\cos t$$

$$g(t)=t\sin t$$

$$U(t)=u(f(t),g(t))$$

Calculate the derivative

$U'(t)$

$$U'(t)=t\sin te^{t^3\cos t\sin t}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t(\sin t+t\cos t)]$$

Solution

We will use the chain rule to calculate the derivative

$$U'(t)=u'_f\cdot f'_t+u'_g\cdot g'_t$$

We have

$$U(t)=u(f(t),g(t))$$

And function u is

$$u(x,y)=e^{xy^2}$$

We put it in the function above and get

$$u(f,g)=e^{fg^2}$$

Hence, the partial derivatives of u are

$$u'_f=g^2e^{fg^2}$$

$$u'_g=2gfe^{fg^2}$$

Also, we calculate the derivatives of f and g

$$f'_t=\cos t -t\sin t$$

$$g'_t=\sin t+t\cos t$$

We put all the derivatives and get

$$U'(t)=u'_f\cdot f'_t+u'_g\cdot g'_t=$$

$$=g^2e^{fg^2}\cdot(\cos t -t\sin t)+2gfe^{fg^2}\cdot (\sin t+t\cos t)=$$

$$=ge^{fg^2}\cdot[g(\cos t -t\sin t)+2f\cdot (\sin t+t\cos t)]=$$

$$=t\sin te^{t\cos t{(t\sin t)}^2}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t\cdot (\sin t+t\cos t)]=$$

$$=t\sin te^{t^3\cos t\sin t}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t(\sin t+t\cos t)]$$

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