Re: Marginal Cost derivative

You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $\displaystyle C\cdot\frac{dC}{dx}$. Now solve for $\displaystyle \frac{dC}{dx}$ by dividing through by $\displaystyle C$.

Re: Marginal Cost derivative

Quote:

Originally Posted by

**MarkFL2** You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $\displaystyle C\cdot\frac{dC}{dx}$. Now solve for $\displaystyle \frac{dC}{dx}$ by dividing through by $\displaystyle C$.

I don't quite understand what you mean...The way I THOUGHT it was supposed to be worked through as is as follows...

$\displaystyle c^2 = x^2 + 100(x)^{1/2} + 50$

$\displaystyle = \frac{d}{dx}C = \frac {d}{dx} x^2 + \frac {d}{dx} 100(x)^{1/2} + \frac {d}{dx} 100$

$\displaystyle = 2C \frac {dC}{dX} = 2x + \frac {50}{x^{1/2}}$

$\displaystyle = \frac {dc}{dx} (C) = x + \frac {25}{x^{1/2}}$

When you say divide through by $\displaystyle C$ what exactly do you mean by that? I thought that once the variables with $\displaystyle \frac {dc}{dx}$ are isolated on one side, that you then factor that out, and then divide the x-valued side, by the C valued side. If I were to do this, wouldn't I end up with this?

$\displaystyle \frac {dC}{dx} = \frac {X}{C} + \frac {25C}{x^{1/2}}$

I didn't think this would be correct because I'm supposed to solve for C. Or now do I plug in x=5, THEN isolate C?

Still a bit lost...

Re: Marginal Cost derivative

You are given:

$\displaystyle C^2=x^2+100x^{\frac{1}{2}}+50$

Implicitly differentiating with respect to $\displaystyle x$, we find:

$\displaystyle 2C\cdot\frac{dC}{dx}=2x+50x^{-\frac{1}{2}}$

$\displaystyle \frac{dC}{dx}=\frac{x+25x^{-\frac{1}{2}}}{C}$

Now, when $\displaystyle x=5$, we have $\displaystyle C=5\sqrt{4\sqrt{5}+3}$ hence:

$\displaystyle \frac{dC}{dx}\left|_{x=5}=\frac{5+5 \sqrt{5}}{5\sqrt{4\sqrt{5}+3}}=\frac{1+\sqrt{5}}{ \sqrt{4\sqrt{5}+3}}\approx0.936349095209$

Re: Marginal Cost derivative

Quote:

Originally Posted by

**MarkFL2**

Now, when $\displaystyle x=5$, we have $\displaystyle C=5\sqrt{4\sqrt{5}+3}$ hence:[/TEX]

Your post makes sense, but I can't see where this value of C is coming from?

Also, wouldn't $\displaystyle 25x^{-1/2} = 25(5)^{-1/2}$,

and not $\displaystyle 25x^{-1/2} = 5(5)^{-1/2}$

Re: Marginal Cost derivative

To find $\displaystyle C$, use:

$\displaystyle C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

$\displaystyle 25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$

Re: Marginal Cost derivative

Quote:

Originally Posted by

**MarkFL2** To find $\displaystyle C$, use:

$\displaystyle C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

$\displaystyle 25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$

Thanks so much for all the help (and patience)!

Re: Marginal Cost derivative

It takes no patience at all to help someone who shows their work/thoughts and wants to understand the problem! :)