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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)

Final Answer

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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