1. ## derivatives

Hello,

I got the following problems and I am not sure how to do them:

fx(x,y) if f(x.y) = 4x-3y+6

fy(x,y) if f(x.y) = 3x^2+2xy-7y^2

and find fx(x.y) and fy(x.y) f(x.y) = square root(2x-y^2)

fxx(x.y) if f(x.y) = 6x-5y+3

and

dz/dx if z = x^3+4x^2y+2y3^3

2. ## Re: derivatives

fsubx is the partial derivative of f with respect to x. So take y as a constant in the f(x,y) expression before differentiating.

Example:
f(x,y)=4x-3y+6
fx(x,y)=4

Similarly, fsuby is the partial derivative with respect to y. Now you will treat x as a constant.
Example:
f(x,y)=3x^2+2xy-7y^2
fy(x,y)=2x-14y

Finally, fxx is the second partial derivative. Just follow the procedure above a second time, with y held constant again.

Hope that helps,
Cheerios

3. ## Re: derivatives

but if y is a constant it would go to 0 when derived no? same for x but you still have x in the final answer.

4. ## Re: derivatives

You must use the product rule. From single-variable calculus:
Given y(x)=cx
y'(x)=c[x(d/dx)]+x[c(d/dx)]=c*1+x*0=c

Look at the term 2xy in the second example function above. The partial derivative with respect to y looks like:
y[2x(d/dy)]+2x[y(d/dy)]=y*0+2x*1=2x