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

$$U'(t)$$

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

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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