# Implicit Differentiation (iii)

• May 5th 2008, 10:13 AM
r_maths
Implicit Differentiation (iii)
• May 5th 2008, 11:03 AM
galactus
The first derivative is correct $\frac{dy}{dx}=\frac{-y^{2}}{x^{2}}$

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:

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

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

Sub in dy/dx from above:

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

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

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

They just used a common denominator and rewrote it. That is the same thing.
• May 5th 2008, 11:18 AM
Soroban
Hello, r_maths!

Quote:

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

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

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

We have: . $\frac{dy}{dx}\:=\:\frac{\text{-}y^2}{x^2}$

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

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

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

• May 5th 2008, 11:29 AM
r_maths
Quote:

Originally Posted by Soroban
Why did you start back there?

(Headbang) I don't know... The book only showed that method.
• May 5th 2008, 11:47 AM
r_maths
Quote:

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

By the way, your last line, shouldn't it equal to this?
$\;\boxed{\frac{2xy^2 + 2y^3}{x^4}}$
• May 5th 2008, 12:54 PM
Moo
Hello,

Quote:

Originally Posted by r_maths
By the way, your last line, shouldn't it equal to this?
$\;\boxed{\frac{2xy^2 + 2y^3}{x^4}}$

Yes, but you could find it by yourself :D
• May 5th 2008, 01:14 PM
r_maths
Quote:

Originally Posted by Moo
Hello,

Yes, but you could find it by yourself :D

:confused: So Soroban is right and the book answer is wrong?
• May 5th 2008, 01:35 PM
Moo
Quote:

Originally Posted by r_maths
:confused: So Soroban is right and the book answer is wrong?

I've done the calculus too, and I find y², not y... :)

So it's 3 against 1 :D