http://img145.imageshack.us/img145/4817/057bp9.png

http://img145.imageshack.us/img145/2858/058fe7.png

What did I do wrong?

Thanks,

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- May 5th 2008, 09:13 AMr_mathsImplicit Differentiation (iii)
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What did I do wrong?

Thanks, - May 5th 2008, 10:03 AMgalactus
The first derivative is correct $\displaystyle \frac{dy}{dx}=\frac{-y^{2}}{x^{2}}$

Now, continue from there by taking the derivative of that. BUT, do not forget to sub in what dy/dx equals. That is where many go astray.

Quotient rule:

$\displaystyle \frac{x^{2}(-2y)\frac{dy}{dx}-(-y^{2})(2x)}{x^{4}}$

$\displaystyle \frac{-2y}{x^{2}}\frac{dy}{dx}+\frac{2y^{2}}{x^{3}}$

Sub in dy/dx from above:

$\displaystyle \frac{-2y}{x^{2}}(\frac{-y^{2}}{x^{2}})+\frac{2y^{2}}{x^{3}}$

$\displaystyle \frac{2y^{2}}{x^{3}}+\frac{2y^{3}}{x^{4}}$

$\displaystyle y''=\frac{2(x+y)y^{2}}{x^{4}}$

They just used a common denominator and rewrote it. That is the same thing. - May 5th 2008, 10:18 AMSoroban
Hello, r_maths!

Quote:

$\displaystyle \frac{1}{x}+\frac{1}{y} \:=\:\frac{1}{\pi}\qquad\text{Find: }\:\frac{dy}{dx}\,\text{ and }\,\frac{d^2y}{dx^2} $

First

$\displaystyle -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{y^2}{x^2}\;\;{\color{blue}\hdots\text{ Right!}}$

Second

$\displaystyle -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\;\;{\color{blue}\hdots \text{Why did you start back there?}}$

Then: . $\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{x^2(\text{-}2y)\left(\frac{dy}{dx}\right) -(\text{-}y^2)(2x)}{(x^2)^2} \;=\; \frac{-2x^2y\cdot\frac{dy}{dx} + 2xy^2}{x^4}$

. . . . $\displaystyle =\;\frac{2xy\left(y - x\!\cdot\!\frac{dy}{dx}\right)}{x^4} \;=\;\frac{2y\left[y - x\left(\text{-}\frac{y^2}{x^2}\right)\right]}{x^3}$

. . . . $\displaystyle = \;\frac{2y\left(y + \frac{y^2}{x}\right)}{x^3} \;=\;\frac{2y(xy + y^2)}{x^4}\;=\;\boxed{\frac{2xy + 2y^3}{x^4}}$

- May 5th 2008, 10:29 AMr_maths
- May 5th 2008, 10:47 AMr_maths
- May 5th 2008, 11:54 AMMoo
- May 5th 2008, 12:14 PMr_maths
- May 5th 2008, 12:35 PMMoo