Evalute partial diff with values of x&y

[Roodles01]

Have worked through to get some sort of answer, but find the evaluation (simplest bit) impossible - wood for trees, I think.

given d2(e^(xy^2)/dy^2 I need to evaluate for x=2, y=0

now using chain rule d(e^(xy^2))/dy = d(e^u)/du du/dy
where u=xy^2 & d(e^u)/du = e^u
=> x(2y) e^(xy^2)
fine for 1st order

moving on
d(x(2y) e^(xy^2))/dy

removing constants
2x(d(ye^(xy^2))/dy

use product rule
d(uv)/dy = v du/dy + u dv/dy
u =e^(xy^2) & v=y

=> 2x(e^(xy^2)(d(y)/dy)+y(d(e^xy^2)))

chain rule again
where u=xy^2 & d(e^u)/du = e^u

2x(ye^(xy^2)(x(d(y^2)/dy)))+e^(xy^2)


which shuld be d2(e^xy^2)/dy2
=> 2x(xy(2y)e^(xy^2)+e^(xy^2)

Now evaluate for x=2, y=0

everything goes to 0 ?

I'v gone through he difficult bits in the book, now the simple stumps me!
Please help before I bang my head against the wall again!

 

[lanedance]

Have worked through to get some sort of answer, but find the evaluation (simplest bit) impossible - wood for trees, I think.

given d2(e^(xy^2)/dy^2 I need to evaluate for x=2, y=0

now using chain rule d(e^(xy^2))/dy = d(e^u)/du du/dy
where u=xy^2 & d(e^u)/du = e^u
=> x(2y) e^(xy^2)
fine for 1st order

$$\frac{\partial}{\partial y}e^{xy^2} = 2xye^{xy^2}$$

looks good so far

moving on
d(x(2y) e^(xy^2))/dy

removing constants
2x(d(ye^(xy^2))/dy

use product rule
d(uv)/dy = v du/dy + u dv/dy
u =e^(xy^2) & v=y

=> 2x(e^(xy^2)(d(y)/dy)+y(d(e^xy^2)))

$$\frac{\partial^2}{\partial^2 y}e^{xy^2} <br /> = \frac{\partial}{\partial y}2xye^{xy^2} <br /> = \left( \frac{\partial}{\partial y}2xy \right) e^{xy^2} +2xy\frac{\partial}{\partial y}e^{xy^2}$$
then you know what the value of the second derivative in the RHS is, as you have already calculated it

below I think you have lost some brackets, that you may want to check

chain rule again
where u=xy^2 & d(e^u)/du = e^u

2x(ye^(xy^2)(x(d(y^2)/dy)))+e^(xy^2)


which shuld be d2(e^xy^2)/dy2
=> 2x(xy(2y)e^(xy^2)+e^(xy^2)

Now evaluate for x=2, y=0

everything goes to 0 ?

I'v gone through he difficult bits in the book, now the simple stumps me!
Please help before I bang my head against the wall again!

if you put the extra backet in and your final expression is
2x(xy(2y)e^(xy^2)) + e^(xy^2)

then knowing e^0=1, this expression is non-zero