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Here are two functions: $f\left(u,v\right)=u^{2}+3v^{2}$

$g\left(x,y\right)=\begin{pmatrix} e^{x}\cos y \\ e^{x}\sin y \end{pmatrix} $

I need to make Jacobian matrix of $f\circ g$. I found derivative of their composition:

$\frac{d\left(f\circ g\right) }{d\left(x,y\right) }=2e^{2x}\cos^{2}{y}+4e^{2x}\sin{y}\cos{y}+6e^{2x}sin^{2}{y} $

How do I put that in Jacobian matrix?

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  • $\begingroup$ What is $\frac{d(f\circ g)}{d(x,y)}$? Try to understand your lesson before getting stuck in exercises. $\endgroup$ Jun 11 '16 at 13:11
  • $\begingroup$ If I understood it I wouldnt post here. $\endgroup$ Jun 11 '16 at 13:15
  • $\begingroup$ In this case you should post a question about what you didn't understand in the lesson ;) $\endgroup$ Jun 11 '16 at 13:16
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Using the chain rule instead: \begin{align*}D(f\circ g)(x,y)& =\color{red}{Df(g(x,y))}\cdot\color{blue}{ Dg(x,y)}\\ & = \color{red}{\begin{pmatrix} 2u&6v \end{pmatrix}\circ(g(x,y))}\cdot \color{blue}{ \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}}\\ & =\color{red}{ \begin{pmatrix} 2e^x\cos y&6e^x\sin y \end{pmatrix}}\cdot \color{blue}{ \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}}\\ \phantom{asd} \\ & = \begin{pmatrix}2e^{2x}\cos^2y + 6e^{2x}\sin^2y & -2e^{2x}\cos y \sin y + 6e^{2x}\sin y\cos y \end{pmatrix}\\ \phantom{asd} \\ & = 2e^{2x}\begin{pmatrix}1 + 2\sin^2y & 2\sin y\cos y \end{pmatrix} \end{align*}

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  • $\begingroup$ But how to put it in matrix? Or the last step is all? $\endgroup$ Jun 11 '16 at 13:25
  • $\begingroup$ The last line is indeed a $1\times 2$ matrix. Just as in @Bye-World's answer $\endgroup$
    – b00n heT
    Jun 11 '16 at 13:26
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    $\begingroup$ @AnaMatijanovic You should get the same matrix whether you use the chain rule or do the composition first and then evaluate the derivative. It'll be instructive to do them both (without looking at b00n's answer) and see if you get the same thing. $\endgroup$
    – user137731
    Jun 11 '16 at 13:28
  • $\begingroup$ Can you explaint two last steps, how did you get 1+2sin^2y ? $\endgroup$ Jun 11 '16 at 13:29
  • $\begingroup$ I just played around with the trigonometric identity $\sin^2 y+\cos^2 y=1$ $\endgroup$
    – b00n heT
    Jun 11 '16 at 13:30
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$$(f\circ g)(x,y) = h(x,y) = e^{2x}\cos^2(y)+3e^{2x}\sin^2(y)$$ Now just build the Jacobian matrix (AKA gradient because $h$ is a scalar-valued function) like normal: $$\pmatrix{\frac{\partial h}{\partial x} & \frac{\partial h}{\partial y}}$$

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