A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Help! Vectorvalued functions.
define \(f:\mathbb{R}^2\rightarrow\mathbb{R}^3\) by \(f(u,v)=(u^25v,ve^{2u},2u\log(1+v^2))\). Suppose \(g:\mathbb{R}^2\rightarrow \mathbb{R}^2\) is of class \(C^1\), \(g(1,2)=(0,0)\) and \[Dg(1,2)=\left[\begin{matrix} 1 & 2 \\ 3 & 5\end{matrix}\right]\]Compute \(D(f\circ g)(1,2)\)
 2 years ago
Help! Vectorvalued functions. define \(f:\mathbb{R}^2\rightarrow\mathbb{R}^3\) by \(f(u,v)=(u^25v,ve^{2u},2u\log(1+v^2))\). Suppose \(g:\mathbb{R}^2\rightarrow \mathbb{R}^2\) is of class \(C^1\), \(g(1,2)=(0,0)\) and \[Dg(1,2)=\left[\begin{matrix} 1 & 2 \\ 3 & 5\end{matrix}\right]\]Compute \(D(f\circ g)(1,2)\)

This Question is Closed

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0so I know that\[Df=\left[\begin{matrix} 2u & 5 \\ 2ve^{2u} & e^{2u} \\ 2 & \frac{2v}{1+v^2}\end{matrix}\right]\]

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0why wouldn't this just be \[Df(1,2)\cdot Dg(1,2)\]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0If you were looking for \(D(f+g)(1,2)\) you could simplify as \(Df(1,2)+Dg(1,2)\), but differentiation is only linear with respect to addition. If you multiplied your functions together, you need to use the chain rule.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.