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!

Related Calculus and Beyond Homework Help News on Phys.org

Math_QED

Homework Helper
Yes, it's correct.

squenshl

Thanks!

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

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving