# Find the second derivative

1. Jul 21, 2012

### menco

1. The problem statement, all variables and given/known data
Find the second derivative of 3(x^2)y+y+x=x^5

2. Relevant equations
Find the first derivative using implicit differentiation.
Find the second by using quotient rule.

3. The attempt at a solution
So I found the first derivative to be

dy/dx = 5(x^4)-6xy-1 / 3(x^2)+1

I get very lost after this trying to find the second derivative using the quotient rule as there is the y variable still in the equation. Any help please?

2. Jul 21, 2012

### ehild

The equation for y' is wrong without parentheses.
Isolate y from the first equation 3(x^2)y+y+x=x^5 and sub into y'.

ehild

3. Jul 21, 2012

### rock.freak667

Put u = 5(x^4)-6xy-1, then get du/dx using implicit differentiation and put v= 3(x^2)+1 and get dv/dx.

Then just put it into your formula for the quotient rule. But in du/dx you will have a term with dy/dx in it (which you know from the first derivative).

4. Jul 21, 2012

### menco

Finding du/dx I get 20(x^3)-6x(dy/dx)-6y and dv/dx I get 6x. Is that correct?

This is what I tried earlier but got very lost trying to sub in dy/dx

Last edited: Jul 21, 2012
5. Jul 21, 2012

### ehild

It is correct now.

ehild

6. Jul 21, 2012

### menco

so then using the quotient rule

dy/dx = (v(du/dx)-u(dv/dx))/(v^2)

= ((3(x^2)+1)*(20(x^3)-(6x((5(x^4)-6xy-1)) / (3(x^2)+1))-6y) - (5(x^4)-6xy-1)* (6x)) / ((3(x^2)+1)^2)

sorry it looks very confusing like that if I knew how to make an image i would.

My calculator gives a result of (2(10(x^3)+27(x^2)(y)+6x-3y) / (3(x^2)+1)^2)
and I have not been able to get close to that answer.

7. Jul 21, 2012

### menco

factoring the above I get a result of

(-60(x^5)+20(x^3)+72(x^2)(y)+12x-6y) / ((3x^2+1)^2)

or

(2(-30(x^5)+10(x^3)+36(x^2)(y)+6x-3y)) / ((3x^2+1)^2)

8. Jul 21, 2012

### HallsofIvy

Staff Emeritus
Finding y' by implicit diferentiation is a good idea and I see no reason not to do the same for the second derivative.

$3x^2y+y+x=x^5$ so
$6xy+ 3x^2y'+ y'+ 1= 5x^4$

Differentiating, implicitely, again,
$6y+ 6xy'+ 6xy'+ 3x^2y''+ y''= 20x^3$
$6y+ 12xy'+ (3x^2+ 1)y''= 20x^3$

You can solve
$3x^2y+y+x=x^5$
for $y= \frac{x^5- x}{3x^2+ 1}$ and
$6xy+ 3x^2y'+ y'+ 1= 5x^4$
for $y'= \frac{5x^4- 6xy+ 1}{3x^2+ 1}$
and put those into $y''= \frac{20x^3- 12xy'- 6y}{3x^2+ 1}$
if you like but my experience is that you seldom have to solve for y'' in terms of x only.