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
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)]
Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!