Stuck getting derivative when can't isolate my variable

[Jon9992]

Hi. This is not a homework assignment. I am working to get an extrema on a graph that involves a bunch of functions and got stuck on one step:

How to get the derivative of:
\frac{dy}{dn} = \frac{nc(a+b)}{nc+a}

I can't get "n" in a place where I recognize how to get the derivative of it. I get stuck here after the first step..
nc(a+b)*(nc+a)^{-1}

or do I use the quotient rule where it ends up:
\frac{[nc(a+b)]'*(nc+a) - nc(a+b)(nc+a)'}{(nc+a)^2}
\frac{c(a+b)*(nc+a) - nc(a+b)(c)}{(nc+a)^2} ??

Can anyone please help?
Thanks very much in advance

 

[Mark44]

Hi. This is not a homework assignment. I am working to get an extrema on a graph that involves a bunch of functions and got stuck on one step:

How to get the derivative of:
\frac{dy}{dn} = \frac{nc(a+b)}{nc+a}

In other words, you want $\frac{d^2 y}{dn^2}$, right? That is, the derivative, with respect to n, of dy/dn.

I can't get "n" in a place where I recognize how to get the derivative of it.

Since you are differentiating with respect to n, then the derivative of n (with respect to n) is just 1. $\frac{d}{dn}(n) = 1$.

I get stuck here after the first step..
nc(a+b)*(nc+a)^{-1}

or do I use the quotient rule where it ends up:
\frac{[nc(a+b)]'*(nc+a) - nc(a+b)(nc+a)'}{(nc+a)^2}
\frac{c(a+b)*(nc+a) - nc(a+b)(c)}{(nc+a)^2} ??

You can use either the product rule (that you started with, above) or the quotient rule - your call. The product rule is usually easier to apply.

Can anyone please help?
Thanks very much in advance

I'm not sure I understood what you were asking. If what I wrote isn't what you're looking for, please clarify your question.

 

[Jon9992]

In other words, you want $\frac{d^2 y}{dn^2}$, right? That is, the derivative, with respect to n, of dy/dn.
Since you are differentiating with respect to n, then the derivative of n (with respect to n) is just 1. $\frac{d}{dn}(n) = 1$.
You can use either the product rule (that you started with, above) or the quotient rule - your call. The product rule is usually easier to apply.

I'm not sure I understood what you were asking. If what I wrote isn't what you're looking for, please clarify your question.

Hi. Thanks. I'm not quite sure what you meant about the d squared y over d n squared. But it sounds like my quotient math above was correct. I wasn't sure if that was right at all.

 

[Jon9992]

In other words, you want $\frac{d^2 y}{dn^2}$, right? That is, the derivative, with respect to n, of dy/dn.
Since you are differentiating with respect to n, then the derivative of n (with respect to n) is just 1. $\frac{d}{dn}(n) = 1$.
You can use either the product rule (that you started with, above) or the quotient rule - your call. The product rule is usually easier to apply.

I'm not sure I understood what you were asking. If what I wrote isn't what you're looking for, please clarify your question.

Actually. I think I know what you meant. I think I wrote that wrong. I meant y=\frac{nc(a+b)}{nc+a}

 

[Mark44]

So all is good, right? You can use either the product rule or the quotient rule for this problem.

I prefer the Leibniz notation for problems like this, rather than the Newton notation. IOW, I prefer d/dn[nc + a] over (nc+a)′, as it isn't clear in the latter notation that you are differentiating with respect to n.