implicit differentiation

**Stuck Man** · May 18th 2012, 04:45 AM

x^-1 +y^-1 = pi^-1

How do I differentiate this? dy/dx should be -(y^2)/x^2.

**Plato** · May 18th 2012, 04:58 AM

Originally Posted by Stuck Man

x^-1 +y^-1 = pi^-1
How do I differentiate this? should be $\frac{dy}{dx}=\frac{-y^2}{x^2}$ .

YES

**Stuck Man** · May 18th 2012, 05:03 AM

That is the answer in the book. I have not obtained it.

**Plato** · May 18th 2012, 05:21 AM

Originally Posted by Stuck Man

That is the answer in the book. I have not obtained it.

Start with $x^{-1}+y^{-1}=\pi^{-1}$
STEP I: $-x^{-2}-y^{-2}y^/=0$

STEP II: Solve for $y^/$

**Stuck Man** · May 18th 2012, 05:38 AM

Thanks. Is the second derivative (2y^3 +2xy^2)/x^4?

**Soroban** · May 18th 2012, 08:06 AM

Hello, Stuck Man!

$\text{Is the second derivative: }\:y'' \:=\:\frac{2y^3 +2xy^2}{x^4}\:?$

Yes . . . but it can be further simplified.

We have: . $y'' \;=\;\frac{2y^2(y + x)}{x^4}$ .[1]

From the original equation: . $\frac{1}{x} + \frac{1}{y} \:=\:\frac{1}{\pi}$

. . we have: . $\frac{y+x}{xy} \:=\:\frac{1}{\pi} \quad\Rightarrow\quad y+x \:=\:\frac{xy}{\pi}$

Substitute into [1]: . $y'' \;=\;\frac{2y^2\left(\frac{xy}{\pi}\right)}{x^4} \;=\;\frac{2y^3}{\pi x^3}$

This is something to watch for . . .
Quite often, the second derivative contains part (or all) of the original equation.

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