Use of derivatives to find coordinates

[Elihu5991]

Homework Statement

Find the coordinates of the point(s) on the following curves where the second derivative is as stated.

Homework Equations

y= \frac{x^3}{12} and \frac{d^{2}y}{dx^{2}} = 1.5

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?

 

[MathematicalPhysicist]

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

 

[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?

 

[Mark44]

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?

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.

 

[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?

 

[HallsofIvy]

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.