Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
2  
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 $y''$

share|improve this answer

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 $f(g(x))' = g'(x) f'(g(x))$)

so for the final answer just add that all together:

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


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

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

share|improve this answer
2  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.