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

#### squenshl

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!

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#### Math_QED

Homework Helper
Yes, it's correct.

#### squenshl

Thanks!

"Finding $(g\circ f)'(6)$"

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