Questions about 2nd Derivatives

[jkh1919]

Hello,

I'm in a calculus class and am currently on a string of problems that are asking me to take the 2nd derivative of a function in order to find its critical numbers, then tell if it is concave upward or downward on a particular interval.

I have not had much trouble finding critical numbers, but 2nd derivatives, have not been so easy. I am going to post a few problems I've been working on for a few days, and hopefully someone can explain to me in a way that I can understand, how to get the derivatives. I'll just go ahead and post the 1st derivatives that I've already come up with and will explain why I am having trouble finding the 2nd derivative.

Here goes:

20x / (9-x^2)^2

I know to use the quotient rule for the whole thing, using the chain rule to find the derivative of the denominator. Here is what I end up with and have no clue how to simplify it properly:

Here are my u and v terms:

u = 20x
v = (9-x^2)^2
uprime = 1
vprime = -4x(9-x^2)

[(9-x^2)^2](1) - (20x)[-4x(9-x^2)] / (9-x^2)^4

I have no clue how to simplify this. My first step would be to multiply -20x * -4x, then take the product (80x), and multiply it by (9-x^2). But this is not what I'm being told. I've been on other forums and spend the entire day yesterday trying to comprehend their answer, but just don't think someone typing it into a forum is going to be enough for me to understand it. But it's the only help I'm getting so I have no choice but to try. So if someone could please explain to me how to do this, I'd greatly appreciate it. I'll post the other problems after I get through this one.

 

[srmichael]

Hello,

I'm in a calculus class and am currently on a string of problems that are asking me to take the 2nd derivative of a function in order to find its critical numbers, then tell if it is concave upward or downward on a particular interval.

I have not had much trouble finding critical numbers, but 2nd derivatives, have not been so easy. I am going to post a few problems I've been working on for a few days, and hopefully someone can explain to me in a way that I can understand, how to get the derivatives. I'll just go ahead and post the 1st derivatives that I've already come up with and will explain why I am having trouble finding the 2nd derivative.

Here goes:

20x / (9-x^2)^2

I know to use the quotient rule for the whole thing, using the chain rule to find the derivative of the denominator. Here is what I end up with and have no clue how to simplify it properly:

Here are my u and v terms:

u = 20x
v = (9-x^2)^2
uprime = 1 ===> NOPE
vprime = -4x(9-x^2)

[(9-x^2)^2](1) - (20x)[-4x(9-x^2)] / (9-x^2)^4

I have no clue how to simplify this. My first step would be to multiply -20x * -4x, then take the product (80x), and multiply it by (9-x^2). But this is not what I'm being told. I've been on other forums and spend the entire day yesterday trying to comprehend their answer, but just don't think someone typing it into a forum is going to be enough for me to understand it. But it's the only help I'm getting so I have no choice but to try. So if someone could please explain to me how to do this, I'd greatly appreciate it. I'll post the other problems after I get through this one.

First, your u' = 1 is incorrect. The derivative of 20x is 20.

Make that change. Then, in the numerator, see what you can factor out of each term. You should see 20(9-x²) as what you can factor out. And you will notice then that (9-x²) can cancel with one of the (9-x²) in the denominator leaving you with (9-x²)^3 in the denominator.

My second derivative is y'' = 20(9+3x²)/(9-x²)^3 or 60(3+x²)/(9-x²)^3

See if you get that.

 

[jkh1919]

First, your u' = 1 is incorrect. The derivative of 20x is 20.

Make that change. Then, in the numerator, see what you can factor out of each term. You should see 20(9-x²) as what you can factor out. And you will notice then that (9-x²) can cancel with one of the (9-x²) in the denominator leaving you with (9-x²)^3 in the denominator.

My second derivative is y'' = 20(9+3x²)/(9-x²)^3 or 60(3+x²)/(9-x²)^3

See if you get that.

Thanks. I realized my mistake shortly after this post. I was having some serious algebraic errors as well.