# implicit derivative of $1=x^3-3xy+y^3$

I'm always having difficulties with rather complicated derivatives, simply because I always make small, stupid mistakes in the process.

Would someone be so kind to help me with the second derivative of $y$ in terms of $y$ and $x$ ?

I did the first: $$\frac{d}{dx}1 = \frac{d}{dx}x^3-\frac{d}{dx}3xy+\frac{d}{dx}y^3 = 3x^2-3y-3x\frac{dy}{dx}+\frac{dy}{dx}(3y^2)$$ $$\Rightarrow \frac{dy}{dx} = \frac{x^2-y}{x-y^2}$$

I got totally lost with the second one though. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{x^2}{x-y^2}-\frac{d}{dx}\frac{y}{x-y^2}=\frac{(2x)(x-y^2)-x^2(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}-\frac{\frac{dy}{dx}(x-y^2)-y(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}$$

And this is basically where I'm loosing it.. I did substitute $\frac{dy}{dx}$ with my first result, but it ended in huge chaos, pretty far from the result wolfram-alpha suggests.

Any help would be appreciated! Also maybe any hints & tricks to avoid such monster-equations (if possible). Because that way I always unnecessarily screw up any test / exam..

Thank you.

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My main advice here would be not to differentiate the solution for $\mathrm dy/\mathrm dx$, but to differentiate the equation again directly and then solve for $\mathrm d^2y/\mathrm dx^2$; that way you avoid differentiating quotients.

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Why the downvote? –  joriki Oct 28 '12 at 10:54
good question.. –  foaly Oct 29 '12 at 13:27

$1=x^3-3xy+y^3$

$0=3x^2-3y-3xy'+3y^2y'$

$y'(3y^2-3x)=3y-3x^2$

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

$y'=\frac{y-x^2}{y^2-x}$

for second derivative from $0=3x^2-3y-3xy'+3y^2y'$ we get

$0=6x-3y'-3y'-3xy''+6yy'y'+3y^2y''$

$6y'-6x-6y(y')^2=y''(3y^2-3x)$

$y''=\frac{2(y'-x-y (y')^2)}{y^2-x}$ replacing that we find for $y'$ finally we get $y''$

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Using the following fact the derivative of A + B is the derivative of A + the derivative of B you can split the problem into smaller parts. Instead of trying to differentiate $x^3-3xy+y^3$ solve the three problems separately:

• differentiate $x^3$ to get $3x^2$... (hint: power rule: $\frac{d}{dx}x^n = n x^{n-1}$)
• differentiate $-3xy$ to get $-3(y + x \frac{dy}{dx})$... (hint: product rule $(fg)' = f'g+fg'$)
• differentiate $y^3$ to get $3 y^2 \frac{dy}{dx}$... (hint: chain rule $f(g(x))' = g'(x) f'(g(x))$)

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

Apply exactly the same idea again (throwing away constants this time)

• $x^2$ gives you $2x$
• $y$ gives $\frac{dy}{dx}$
• $x \frac{dy}{dx}$ gives $\frac{dy}{dx} + x \frac{d^2y}{dx^2}$
• $y^2 \frac{dy}{dx}$ gives $y^2 \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2$

and add it back together (with the right constants) to get

$$3\left(2x - \frac{dy}{dx} - \frac{dy}{dx} - x \frac{d^2y}{dx^2} + y^2 \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2\right)$$

simplify

$$3\left(2x - 2 \frac{dy}{dx} + (y^2 - x) \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2\right)$$

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This does not address the question. –  wj32 Oct 28 '12 at 10:24
implicit derivative of 1=x3−3xy+y3 –  sperners lemma Oct 28 '12 at 10:25
that's the title, if you read the question you would have seen that that's pretty much what I did, and instead was struggling with the second derivative. I suppose what you actually wanted to say is I should have written 'second' in the title. I will next time. –  foaly Oct 28 '12 at 10:28
@foaly, I didn the second derivative now –  sperners lemma Oct 28 '12 at 10:33
+1 for balance... –  wj32 Oct 28 '12 at 10:42