# Use of derivatives to find coordinates

1. Feb 27, 2013

### Elihu5991

1. The problem statement, all variables and given/known data
Find the coordinates of the point(s) on the following curves where the second derivative is as stated.

2. Relevant equations
$y= \frac{x^3}{12} and \frac{d^{2}y}{dx^{2}} = 1.5$

3. The attempt at a solution
I'm used to working with the first derivative. Would I need to use integration to convert the function to the first derivative then treat it like a usual gradient - derivative = gradient? If I do need to continue to work with the second derivative, could I receive some hints on how?

2. Feb 27, 2013

### MathematicalPhysicist

Well you just need to calculate y'' explicitly and then find x such that y''=1.5.

3. Feb 27, 2013

### Elihu5991

Yeah it does seem to be the case. Though I'm getting really large numbers. I'm using the quotient rule twice to differentiate. Is that correct?

4. Feb 27, 2013

### Staff: Mentor

You're making this problem much more difficult than it actually is. For your function you should NOT use the quotient rule. It's not wrong to do so, but it's a more complicated method that is more likely to lead to errors.

You should never use the quotient rule to differentiate a quotient with a constant in the denominator. Instead write the function as (1/k) * f(x) and use the constant multiple rule.

d/dx(1/k * f(x)) = 1/k * f'(x)

The same thinking holds for functions of the form y = k * f(x). Although this is a product, the natural tendency would be to use the product rule. The easier rule to use would be the constant multiple rule here as well.

5. Feb 28, 2013

### Elihu5991

Yeah that's true. I'm known for unintentionally doing so.

From what I'm gathering I write it like this: $\frac{x^{3}}{12}$ to $x^{3}12^{-1}$ and then go from their by normally differentiating twice over to get the second derivative - consequently finding x for the coordinate to substitute into the original to get the y for the final coordinates?

6. Feb 28, 2013

### HallsofIvy

Staff Emeritus
Why in the world write (1/12) as $12^{-1}$? Surely you know that (Cf(x))'= Cf'(x) for any constant C? Just differentiate $x^3$ twice and divide that by 12.