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Multivariable calculus

Unit 2: Lesson 5

Multivariable chain rule
Math
Multivariable calculus
Derivatives of multivariable functions
Multivariable chain rule
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Multivariable chain rule intro

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Problem

Suppose we have a vector-valued function g, left parenthesis, t, right parenthesis and a scalar function f, left parenthesis, x, comma, y, comma, z, right parenthesis. Let h, left parenthesis, t, right parenthesis, equals, f, left parenthesis, g, left parenthesis, t, right parenthesis, right parenthesis.
We know:
g(1)=(0,2,4)g′(1)=(−1,6,2)∇f(0,2,4)=(3,5,1)\begin{aligned} &g(1) = (0, 2, 4) \\ \\ &g'(1) = (-1, 6, 2) \\ \\ &\nabla f(0, 2, 4) = (3, 5, 1) \end{aligned}
Evaluate start fraction, d, h, divided by, d, t, end fraction at t, equals, 1.
h, prime, left parenthesis, 1, right parenthesis, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
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