Math Help - Implicit Differentiation (iii)

1. Implicit Differentiation (iii)

What did I do wrong?

Thanks,

2. 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.

3. Hello, r_maths!

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

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

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

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

4. Originally Posted by Soroban
Why did you start back there?
I don't know... The book only showed that method.

5. 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?
$\;\boxed{\frac{2xy^2 + 2y^3}{x^4}}$

6. Hello,

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

7. Originally Posted by Moo
Hello,

Yes, but you could find it by yourself
So Soroban is right and the book answer is wrong?

8. Originally Posted by r_maths
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