Question

how is f x y = (-y sin xy)(x) = -xy sin xy ?

Answer 1

how to solve this second order partial derivative? when function f(x,y) =Sin xy.....??? i have differentiated w.r.t. x and got
\[-y ^{2} Sin xy\]

but to find 2nd order part. derv. i have differentiated this w.r.t. y and i have got
\[-xy Sin xy -\cos xy\]

is this correct?

Answer 2

is it?

Answer 3

When you take a partial w.r.t. x you hold y as a constant, so:
\[D_x=-ysin(xy)-xy^2\cos(xy)\]

Answer 4

plz hold on..i'll type how i had solved...

Answer 5

function is 'Sin xy'
part.derv. wrt x=y cos xy -------eq. 1
part.derv. of eq .1 wrt x= \[-y ^{2} \sin xy\]
and part.derv of eq 1 wrt y is = \[-xy \sin xy-\cos xy \] (using uv rule)

Answer 6

is the part.derv. of eq 1 wrt y correct?

Answer 7

to find part.derv. of eq 1 wrt 'y' :-
\[u= y \cos xy & v=\cos xy\]
then, \[{y(- \sin xy)x}-{\cos xy .1}\]
and thats how i got
−xysinxy−cosxy

Answer 8

u=y and v=cos xy

Answer 9

are you getting the same answer?

Answer 10

u der?

Answer 11

I thought your function was -xysin(xy)......? your partial w.r.t. x is correct what are you trying to find out ?
\[f_x(x,y), f_y(x,y), or, f_{xy}(x,y)\]

Answer 12

the actual function is f(x,y)=Sin xy
\[f_{x} (x,y)= y Cos xy \] ,
\[f _{xx}= - y ^{2} Sin xy \] and \[f _{xy}(x,y)= -xy Sin xy+\cos xy\]

Answer 13

is this correct???

Answer 14

Yes, thats correct

Answer 15

yeah i got the answer

Answer 16

yeah i got the answer

Answer 17

cool