2nd Derivative Implicit

**AZach** · October 25th 2012, 02:18 PM

Find $\frac {d^2y}{dx^2}$

$e^{2y}+x = y$

I didn't get very far after differentiating with respect to x; here's my step 1): $2\frac{dy}{dx}*e^{2y}\frac{dy}{dx} + 1 = \frac{dy}{dx}$ .

I have several questions. Can I multiply $\frac{dy}{dx}$ by another to get $\frac {d^2y}{dx^2}$ ?

If not, I'll show the steps I went through with a brief explanation: I moved a dy/dx and the +1 to different sides and then factored out a dy/dx:
2) $\frac{dy}{dx}(2e^{2y}-1) = -1$

Solved for dy/dx
3) $\frac{dy}{dx}= \frac{-1}{2e^{2y}-1}$

Differentiated again with respect to x using quotient rule / chain rule.
4) $\frac {d^2y}{dx^2} = {\frac{(-2e^{2y}\frac{dy}{dx}*2\frac{dy}{dx})*-1}{(2e^{2y}-1)^2}$

Simplified the top by distributing the negative one and multiplying the dy/dx terms (if that's legal).
5) $\frac {d^2y}{dx^2} = {\frac{4e^{2y}\frac{dy}{dx}}{(2e^{2y}-1)^2}$

Substituted dy/dx back into the game.
6) $\frac {d^2y}{dx^2} = {\frac{4e^{2y}*\frac{-1}{2e^{2y}-1}}{(2e^{2y}-1)^2}$

Multiplied lowest denominator across the middle denominator.
7) $\frac {d^2y}{dx^2} = {\frac{4e^{2y}}{(2e^{2y}-1)^3}$

**skeeter** · October 25th 2012, 03:02 PM

$\frac{d}{dx} \left(e^{2y} + x = y \right)$

$e^{2y} \cdot 2\frac{dy}{dx} + 1 = \frac{dy}{dx}$

$1 = \frac{dy}{dx} - e^{2y} \cdot 2\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{1 - 2e^{2y}}$

$\frac{d}{dx} \left[\frac{dy}{dx} = (1 - 2e^{2y})^{-1} }\right]$

$\frac{d^2y}{dx^2} = -(1 - 2e^{2y})^{-2} \cdot (-4e^{2y}) \cdot \frac{dy}{dx}$

$\frac{d^2y}{dx^2} = \frac{4e^{2y}}{(1 - 2e^{2y})^2} \cdot \frac{1}{1 - 2e^{2y}} = \frac{4e^{2y}}{(1 - 2e^{2y})^3}$

I almost agree ...

**AZach** · October 25th 2012, 06:20 PM

I didn't realize I had almost gotten that right. Thanks for showing me your work, Skeeter.

**AZach** · October 27th 2012, 02:57 PM

I have another problem of a similar nature to add to this. $sin(x) + x^{2}y = 1$

First derivative with respect to x: 1) $cos(x) + 2xy+x^{2}*\frac{dy}{dx} = 0$

Solve for dy/dx: 2) $\frac{dy}{dx} = \frac{-cos(x)-2xy}{x^{2}}$

Now things get hairy taking second d: 3) $\frac{(sin(x)-2y-2x\frac{dy}{dx})x^{2}-(-cos(x)-2xy)2x}{x^{4}}$

Substituting dy/dx back into the game: 4) $\frac{(sin(x)-2y-2x(\frac{-cos(x)-2xy}{x^{2}}))x^{2}-(-cos(x)-2xy)2x}{x^{4}}$

step 5), $\frac{(sin(x)-2y+(\frac{(2x)cos(x)-4x^{2}y}{x^{2}}))x^{2}+(2x)cos(x)-4x^{2}y}{x^{4}}$

And then, $\frac{(sin(x)-2y+{(2x)cos(x)-4x^{2}y})x^{2}+(2x)cos(x)-4x^{2}y}{x^{6}}$

$\frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)-4x^{4}y+(2x)cos(x)+4x^{2}y}{x^{6}}$

I get stuck here now. I can't factor out the 2. $\frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)+(2x)cos(x)}{x^{6}}$

**HallsofIvy** · October 27th 2012, 04:37 PM

Originally Posted by skeeter

$\frac{d}{dx} \left(e^{2y} + x = y \right)$

$e^{2y} \cdot 2\frac{dy}{dx} + 1 = \frac{dy}{dx}$

$1 = \frac{dy}{dx} - e^{2y} \cdot 2\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{1 - 2e^{2y}}$

Personally, I would not solve for dy/dx here. From
$2e^{2y}\frac{dy}{dx}+ 1= \frac{dy}{dx}$
use "implicit differentiation" again
$4e^{2y}\frac{dy}{dx}+ 2e^{2y}\frac{d^2y}{dx^2}= \frac{d^2y}{dx^2}$
Then $\frac{d^2y}{dx^2}= \frac{4e^{2y}\frac{dy}{dx}}{1- 2e^{2y}}$
And, if you are required to write it in terms of x only, go back and and solve for $\frac{dy}{dx}$

$\frac{d}{dx} \left[\frac{dy}{dx} = (1 - 2e^{2y})^{-1} }\right]$

$\frac{d^2y}{dx^2} = -(1 - 2e^{2y})^{-2} \cdot (-4e^{2y}) \cdot \frac{dy}{dx}$

$\frac{d^2y}{dx^2} = \frac{4e^{2y}}{(1 - 2e^{2y})^2} \cdot \frac{1}{1 - 2e^{2y}} = \frac{4e^{2y}}{(1 - 2e^{2y})^3}$

I almost agree ...

**HallsofIvy** · October 27th 2012, 04:44 PM

Originally Posted by AZach

I have another problem of a similar nature to add to this. $sin(x) + x^{2}y = 1$

First derivative with respect to x: 1) $cos(x) + 2xy+x^{2}*\frac{dy}{dx} = 0$

Differentiating here, $-sin(x)+ 2y+ 2xy'+ 2xy'+ x^2+ x^2y''= 0$ , y''= \frac{sin(x)- 2y- 4xy'}{x^2}[/tex]

Solve for dy/dx: 2) $\frac{dy}{dx} = \frac{-cos(x)-2xy}{x^{2}}$

Now things get hairy taking second d: 3) $\frac{(sin(x)-2y-2x\frac{dy}{dx})x^{2}-(-cos(x)-2xy)2x}{x^{4}}$

Substituting dy/dx back into the game: 4) $\frac{(sin(x)-2y-2x(\frac{-cos(x)-2xy}{x^{2}}))x^{2}-(-cos(x)-2xy)2x}{x^{4}}$

step 5), $\frac{(sin(x)-2y+(\frac{(2x)cos(x)-4x^{2}y}{x^{2}}))x^{2}+(2x)cos(x)-4x^{2}y}{x^{4}}$

And then, $\frac{(sin(x)-2y+{(2x)cos(x)-4x^{2}y})x^{2}+(2x)cos(x)-4x^{2}y}{x^{6}}$

$\frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)-4x^{4}y+(2x)cos(x)+4x^{2}y}{x^{6}}$

I get stuck here now. I can't factor out the 2. $\frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)+(2x)cos(x)}{x^{6}}$

**AZach** · October 27th 2012, 05:05 PM

Originally Posted by HallsofIvy

Differentiating here, $-sin(x)+ 2y+ 2xy'+ 2xy'+ x^2+ x^2y''= 0$ , $y''= \frac{sin(x)- 2y- 4xy'}{x^2}$

I see you did the product rule, but I'm confused. You got $2xy'+x^2$ which I think is the result of differentiating $x^{2}y'$ , but then there's also an $x^{2}y''$ at the end of the left side of the expression. When you differentiate $\frac{dy}{dx}$ using the product rule does that term go to 1? Wait, is that single $x^2$ term before solving for $y''$ a typo? I don't see it moved over to the right hand side.

Thanks!

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