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
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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