## anonymous 5 years ago how is f x y = (-y sin xy)(x) = -xy sin xy ?

1. anonymous

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?

2. anonymous

is it?

3. anonymous

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

4. anonymous

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

5. anonymous

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)

6. anonymous

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

7. anonymous

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

8. anonymous

u=y and v=cos xy

9. anonymous

are you getting the same answer?

10. anonymous

u der?

11. anonymous

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

12. anonymous

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

13. anonymous

is this correct???

14. anonymous

Yes, thats correct

15. anonymous

16. anonymous