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Finding ##(g\circ f)'(6)##

  • Thread starter squenshl
  • Start date
Homework Statement
Suppose that ##\alpha,\beta: \mathbb{R}\to \mathbb{R}## and ##g: \mathbb{R}^2\to \mathbb{R}## are differentiable functions and ##f: \mathbb{R}\to \mathbb{R}^2## is the function defined by ##f(t) = (\alpha(t),\beta(t))##. Suppose further that
$$f(6) = (10,-10), \quad \alpha'(6) = 4, \quad \beta'(6) = -3, \quad g_x(10,-10) = -1, \quad \text{and} \quad g_y(10,-10) = -2.$$
Then ##(g\circ f)'(6)## equals to
1. ##24##.
2. ##-24##.
3. ##2##.
4. ##0##.
5. ##-2##.
Homework Equations
Multivariate chain rule formula.
The solution is 3: It's just ##(g\circ f)'(6) = (-1,-2)\cdot (4,-3) = (-1\times 4)+((-2)\times (-3)) = -4+6 = 2## using the multi-variate chain rule and the dot product.

Is this correct and if not how do I go about doing it?
Thanks!
 

Math_QED

Science Advisor
Homework Helper
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Yes, it's correct.
 
Thanks!
 

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