Partial Derivatives of a Function

C

CatchThis2

Junior Member
Joined
Feb 6, 2010
Messages
96
Find the partial derivative of the function : F(x,y)= -9x+3y / -7x-2y

Find Fx(x,y)
Fy(x,y)


I got this answer but I think it is incorrect: Fx(x,y)=69y/(-7x-2y)^2
Fy(x,y)= -7x/(-7x-2y)^2
 
CatchThis2 said:
Find the partial derivative of the function : F(x,y)= -9x+3y / -7x-2y

Find Fx(x,y)
Fy(x,y)


I got this answer but I think it is incorrect: Fx(x,y)=69y/(-7x-2y)^2
Fy(x,y)= -7x/(-7x-2y)^2
Click to expand...

Assuming:

\(\displaystyle F(x,y) = \frac{-9x + 3y}{-7x - 2y}\)

I get:

\(\displaystyle F_x(x,y) = \frac{39y}{(7x+2y)^2}\)

\(\displaystyle F_y(x,y) = \frac{39x}{(7x+2y)^2}\)
 
Let Z=2e^x^4^y^4

Then z/x=

z/y=


For the first question I got 8e^x^3(y^4) but I don't think that is right?
 
CatchThis2 said:
Let Z=2e^x^4^y^4

Then z/x= <<<< What does that mean??

z/y=<<<< What does that mean??


For the first question I got 8e^x^3(y^4) but I don't think that is right?
Click to expand...
 
It means the partial derivative of z divided by x and the partial derivative of z over y
 
\(\displaystyle Let \ z \ = \ 2e^{x^{4^{y^{4}}.\)

\(\displaystyle What \ is \ this? \ Are \ you \ all \ right?\)
 
For such problems take ln on both sides. Remember that it is loge() So there may be loge(10) terms when you multiply. When you are taking Fx assume y term to be constant and x in case of fx. I have solved Fx below.
Since we are diff wrt x, assume 4^y^4 to be a constant c
z=2e^x^c
ln(z/2) =x^c
Differentiating on both sides,
2/z(dz/dx)=c*x^c-1
dz/dx=(zc/2)x^c-1
 
You must log in or register to reply here.
Top