3x^2y / (e^xy+2)
Fx: 6xy(e^xy+2) - ye^xy(3x^2y) / (e^xy+2)^2 <--- this is where i got to, but stopped since it was right first of all, and i just dont understand the partial derivative of e^xy in terms of x.
(im doing the partial derivative of x)
3x^2y / (e^xy+2)
Fx: 6xy(e^xy+2) - ye^xy(3x^2y) / (e^xy+2)^2 <--- this is where i got to, but stopped since it was right first of all, and i just dont understand the partial derivative of e^xy in terms of x.
(im doing the partial derivative of x)
For partial derivative just treat any variables other than the one you are differentiating with respect to as constants.
so the x partial of $\displaystyle \frac{6y}{x+y}\text{ is no different than the x partial of }\frac{6c}{x+c}$ but we'll replace c with y.
That should be a derivative you can do. It's just $\displaystyle \frac{-6c}{(x+c)^2} \text{which becomes again }\frac {-6y}{(x+y)^2}$
You don't usually explicitly put the c in there. You just learn to treat the other variables as constants.
The y partial of this expression is a little trickier but not that much more. See if you can figure that one out.
its the first one you have stated, and I forgot to say that this is asking for the second partial derivative.
The answer is 12 on top i just dont understand how
and isnt the first, -6y/(x+y)^2
the 2nd derv answer is 12/(x+y)^3
Yes it is -6y/(x+y)^2
If you know that you can write 6y/x+y as 6y(x+y)^-1
Then when you differentiate, bring the power down to the front and reduce the power by -1, so it becomes
-6y(x+y)^-2
Then you can bring (x+y)^-2 underneath again to make it -6y/(x+y)^2
The second derivative is the same process again, try it out, can you see it now?
yessir you are correct, it is negative. I gotta stop being so sloppy.
$\displaystyle f_x[x]=\frac {-6y}{(x+y)^2}=-6y(x+y)^{-2}$
$\displaystyle f_{xx}[x]=(-2)(-6y)(x+y)^{-3}=12y(x+y)^{-3}=\frac{12y}{(x+y)^3}$
so you want the 2nd derivative for that other term now?
$\displaystyle f[x,y]=\frac{3x^{2}y}{e^{xy}+2}$
basically you just use the quotient rule treating y as a constant. If you could figure the other one out you should be able to figure this one out.
hmmm, I can't get white text to work atm. Yell back if you want to check your answer.