# Implicit differentiation

• Nov 14th 2012, 06:58 AM
Frozenboy11
Implicit differentiation
Here my question , for example x^2*y^2=0 when y=x=1, find d/dx, can I make the y^2 = x^2 and therefor change the equation to x^4=0 and differentiate for my x?
• Nov 14th 2012, 07:04 AM
Plato
Re: Implicit differentiation
Quote:

Originally Posted by Frozenboy11
Here my question , for example x^2*y^2=0 when y=x=1, find d/dx, can I make the y^2 = x^2 and therefor change the equation to x^4=0 and differentiate for my x?

That is an impossible example.
$x^2y^2=0$ when $y=x=1$ does not exist.
• Nov 14th 2012, 07:06 AM
Frozenboy11
Re: Implicit differentiation
Ok my bad let said it is equal to some number so can I let y^2=x^2? When x=y=1?
• Nov 14th 2012, 07:20 AM
Plato
Re: Implicit differentiation
Quote:

Originally Posted by Frozenboy11
Ok my bad let said it is equal to some number so can I let y^2=x^2? When x=y=1?

If $y^2=x^2$ then $2yy^{\prime}=2x$ or $y^{\prime}=\frac{x}{y}$.

SO?
• Nov 14th 2012, 07:30 AM
Frozenboy11
Re: Implicit differentiation
Quote:

Originally Posted by Plato
If $y^2=x^2$ then $2yy^{\prime}=2x$ or $y^{\prime}=\frac{x}{y}$.

SO?

Here the full question
• Nov 14th 2012, 08:16 AM
HallsofIvy
Re: Implicit differentiation
Yes, that is wrong. "At x= y= 1" does not mean that you can set x= y before taking the derivative. You have to take the derivative, with respect to y as (1, y) approaches (1, 1). So, because this is the partial derivative with respect to 1, you can set x= 1, making $f(x, y)= ln((1+2y)^2+ (2- y)^2+ 4y^2)$. Differentiate that and then set y= 1.