A multivariate calculus function?

Question

f(x,y) = (sin(x)sin(y))/xy what is the second partial derivative of this function. I want only the second partial derivative w.r.t x, differentiate w.r.t x first.

Answer 1

Let df/dx and d^2f/x^2 denote the first and second partials with respect to x of f(x,y)
When we take partials of f with respect to x we hold all terms in y constant
df/dx = [sin(y)/y][(x cos(x) - sin(x))/x^2]

= [sin(y)/y][cos(x)/x - sin(x)/x^2]

Now we compute the first partials of cos(x)/x and sin(x)/x^2

d/dx cos(x)/x = [-x sin(x) - cos(x)]/x^2

d/dx sin(x)/x^2 = [x^2 cos(x) - 2x sin(x)]/x^4

Now we compute d^2f/dx^2

d^2f/dx^2 = [sin(y)/y]d/dx[cos(x)/x - sin(x)/x^2]

= [sin(y)/y][ d/dx cos(x)/x - d/dx sin(x)/x^2]

Now substitute for the partials in the second pair of brackets and simplify

Answer 2

Assume y is a constant and twice differentiate the function w.r.t. x just like you did in calculus 1. If you don't know how to do that, then that is a clue that you are in over your head in whatever class you are taking. Good luck.

Answer 3

To differentiate with respect to x, you treat y and functions of y as constants. The first derivative of f(x,y) requires the quotient rule, (f/g)' = (gf' - fg')/g^2, and you get (xy*cos(x)sin(y) - sin(x)sin(y)*y)/(x^2*y^2). The second derivative requires a second application of the quotient rule. It gives you ((x^2*y^2)*(sin(y)y(cos(x) - sin(x)x) - (xy*cos(x)sin(y) - sin(x)sin(y)*y)*2x*y^2)/(x^4*y ^4). Nobody ever promised you a bed of roses.

Answer 4

the two on the prompt at the instant are no further common, yet calculus III should not be a prerequesite for differential equations (so which you would be waiting to now not want one for the different), so as long as you may desire to shield the mathematics you would be effective.