Math Analysis

[SpanishTeacher]

Hi everyone! Can you help me with this question? Thanks!

 

[chandra21]

can not see a letter

 

[stapel]

Hi everyone! Can you help me with this question? Thanks!

View attachment 2788

To other users: The text in the above (tiny!) graphic is as follows:

In each part, consider the function \(\displaystyle f\) that is defined and calculate \(\displaystyle \left(Df\right)_p (h)\), \(\displaystyle \left(Df\right)^2_p (h,k)\), and \(\displaystyle \left(Df\right)^3_p (h,k,l)\) for:

\(\displaystyle p\, =\, \left(\begin{array}{c}x\\y\end{array}\right)\, \mbox{ and }\, p\, =\, \left(\begin{array}{c}1\\0\end{array}\right)\)​

where

\(\displaystyle h\, =\, \left(\begin{array}{c}h_1\\h_1\end{array}\right),\, k\, =\, \left(\begin{array}{c}k_1\\k_1\end{array}\right),\, \mbox{ and }\, l\, =\, \left(\begin{array}{c}l_1\\l_1\end{array}\right)\)​

are points in \(\displaystyle \mathbb{R}^2\).

a) \(\displaystyle f\, :\, \mathbb{R}^2\, \rightarrow\, \mathbb{R}\) is defined by \(\displaystyle f\, \left(\begin{array}{c}x\\y\end{array}\right)\, =\, x^3\, +\, 2xy\, -\, e^y\)

b) \(\displaystyle f\, :\, \mathbb{R}^2\, \rightarrow\, \mathbb{R}^2\) is defined by \(\displaystyle f\, \left(\begin{array}{c}x\\y\end{array}\right)\, =\, \left(\begin{array}{c}x^3\, +\, 2xy\, -\, e^y\\xy\, =\, 1\end{array}\right)\)

 

[HallsofIvy]

So you are given two functions, one from \(\displaystyle R^2\) to \(\displaystyle R\) the other from \(\displaystyle R^2\) to \(\displaystyle R^2\) and are asked to find the first, second, and third derivatives. Okay, what is your difficulty? Do you know how to find derivatives of higher dimension functions?

A derivative is a linear function that approximates the original function. A linear function from \(\displaystyle R^2\) to\(\displaystyle R\) can be thought of as a vector <a, b> such that the dot product, <a, b>.<x, y>, gives the value of the function. It shouldn't be too surprising that this is given by \(\displaystyle <a, b>= \left<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right>\)

Similarly, the derivative of a function from \(\displaystyle R^2\) to \(\displaystyle R^2\) can be thought of as a two by two matrix:
\(\displaystyle \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\)