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Multivariable Chain Rule – Exercise 3313

Exercise

Given

u(x,y,z)=x^2+y^2+xz

f(t)=\sin (4t)

g(t)=e^{-t}

h(t)=t^3

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

Calculate the derivative

U'(t)

Final Answer

U'(t)=(2\sin (4t)+t^3)\cdot 4\cos (4t)-2e^{-2t}+3t^2\sin(4t)

Solution

We will use the chain rule to calculate the derivative

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

We have

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

And function u is

u(x,y,z)=x^2+y^2+xz

Hence, we get

u(f,g,h)=f^2+g^2+fh

We calculate the partial derivatives of u.

u'_f=2f+h

u'_g=2g

u'_h=f

And the derivatives of f,g and h.

f'_t=4\cos (4t)

g'_t=-e^{-t}

h'_t=3t^2

We put the derivatives in U’.

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

=(2f+h)\cdot 4\cos (4t)+2g\cdot (-e^{-t})+f\cdot 3t^2=

=(2\sin (4t)+t^3)\cdot 4\cos (4t)+2e^{-t}\cdot (-e^{-t})+\sin(4t)\cdot 3t^2=

=(2\sin (4t)+t^3)\cdot 4\cos (4t)-2e^{-2t}+3t^2\sin(4t)

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