Calculus experts please HELP A BROTHA OUT!!!?

Question

What is the second-order partial derivative with respect to x and y of the function: f(x,y)=x^2sin(3xy)?

<<>>
THANKS GUYS!!! :)

THANK U ALL!!!

Answer 1

I take it you want what I will call:

partial d/dx of partial d/dy of f(x, y) = x^2 sin (3 x y). (I like to separate the variables out a bit, so that I can more easily see and deal with them.)

There's a theorem that for a continuous function, the order in which you do the partial derivatives doesn't matter. So I'll actually invert the order, and do the partial d/dx first. Let's see where that gets us:

partial d/dx [x^2 sin (3 x y)] = 2 x sin (3xy) + x^2 3 y cos (3xy).

Now do a partial d/dy of THAT. Why not see if you can do it, step by step, yourself. (I think your teacher will want to see those intermediate steps, rather than a suspicious leap to the final result.)

Here, to help you, but without a cast-iron guarantee, is what I ultimately find:

partial d/dy of partial d/dx f(x, y) = 9 x^2 cos (3xy) - 9 x^3 y sin (3xy).

I hope this helps.

Live long and prosper.

Answer 2

When taking the derivative of a multivariable function with respect to x, treat y as a constant. Vice versa for y; treat x as a constant.

Try starting from there.

Answer 3

to the respect of x...
f(x) = uv
f'x = u'v + v'u

u = x^2 u'= 2x
v = sin 3xy v' =

f'(x) = 2x sin (3xy) + 3yx^2.cos(3xy).
same thing,
f'' (x) = 2 sin (3xy) + 2x.cos(3xy).3y + 6yx cos(3xy) + 3yx^2 (-sin(3xy) 3y)
f''(x) = 2sin(3xy) + 6xycos(3xy) + 6xycos(3xy) - 9y^2x^2sin(3xy)
f''(x) = (2-9y^2x^2) sin(3xy) + 12xy cos(3xy)

to the respect of y...
f (y) = x^2 sin (3xy)

f'(y) = x^2 cos(3xy) 3x
f'(y) - 3x^3 cos (3xy)

f''(y) = 3x^3 (- sin(3xy)) 3x
f''(y) = - 27x^4 sin(3xy)

Answer 4

i wish i was a derivative so i could lie tangent to your curves

Answer 5

http://img247.imageshack.us/img247/1953/untitled504dj6.jpg