- #1
Roodles01
- 128
- 0
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 throught he difficult bits in the book, now the simple stumps me!
Please help before I bang my head against the wall again!
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 throught he difficult bits in the book, now the simple stumps me!
Please help before I bang my head against the wall again!