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

Show final answer

Solution

We will use the chain rule to calculate the derivative

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

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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