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

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

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- May 18th 2012, 04:45 AMStuck Manimplicit differentiation
x^-1 +y^-1 = pi^-1

How do I differentiate this? dy/dx should be -(y^2)/x^2. - May 18th 2012, 04:58 AMPlatoRe: implicit differentiation
- May 18th 2012, 05:03 AMStuck ManRe: implicit differentiation
That is the answer in the book. I have not obtained it.

- May 18th 2012, 05:21 AMPlatoRe: implicit differentiation
- May 18th 2012, 05:38 AMStuck ManRe: implicit differentiation
Thanks. Is the second derivative (2y^3 +2xy^2)/x^4?

- May 18th 2012, 08:06 AMSorobanRe: implicit differentiation
Hello, Stuck Man!

Quote:

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

Yes . . . but it can be further simplified.

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

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

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

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

This is something to watch for . . .

Quite often, the second derivative contains part (or all) of the original equation.