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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you take a partial w.r.t. x you hold y as a constant, so: \[D_x=ysin(xy)xy^2\cos(xy)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plz hold on..i'll type how i had solved...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0function 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the part.derv. of eq 1 wrt y correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you getting the same answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought your function was xysin(xy)......? your partial w.r.t. x is correct what are you trying to find out ? \[f_x(x,y), f_y(x,y), or, f_{xy}(x,y)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the actual function is f(x,y)=Sin xy \[f_{x} (x,y)= y Cos xy \] , \[f _{xx}=  y ^{2} Sin xy \] and \[f _{xy}(x,y)= xy Sin xy+\cos xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i got the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i got the answer
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