Implicit Derivative of this Function

[k_squared]

[Solved]Implicit Derivative of this Function

Homework Statement

Find the implicit Derivative of this function:
1/x+1/y=1

Homework Equations

Chain Rule, Quotient Rule ... ?

The Attempt at a Solution

1/x-(y'/y^2)=1
(y'/y^2)=1-1/x
y'=1/y^2-y^2/x

The answer seems to be (-y^2/x^2). I have no idea how that happened. Any help would be most appreciated...

 

[dreit]

It doesn't look like u took the derivative of the (1/x) from the work provided

 

[dreit]

Also you may want to try to rewrite your function as x^(-1)+y^(-1)=1

 

[k_squared]

My new work goes something like

-1/x^2+y'/y^2=1
y'/y^2=1-1/x^2

y' = y^2((x^2+1)/x^2)
y' = y^2-y^2/x^2


Uh... Could you tell me what I'm doing wrong (Ie, is this an algebra mistake or is my calc itself off)? Again, any help is appreciated...
Forgot to realize the derivative of 1 is zero... sorry...

 

[HallsofIvy]

My new work goes something like

-1/x^2+y'/y^2=1

You original function was given by 1/x+ 1/y= 1
You should have a "-" on both x and y and the derivative of 1 is 0.

y'/y^2=1-1/x^2

y' = y^2((x^2+1)/x^2)
y' = y^2-y^2/x^2


Uh... Could you tell me what I'm doing wrong (Ie, is this an algebra mistake or is my calc itself off)? Again, any help is appreciated...
Forgot to realize the derivative of 1 is zero... sorry...