Average Cost

[johnjones]

Can someone give me a hint for this question:

Given that C(10) = 120 and C'(10) =8, find d/dx[C(x)/x)] for x = 10.
Do I use the quotient rule? I don't know the function of C(x)? Thx.

 

[Gene]

Yup, that will give an equation in C(x),C'(x) and x and you know what they are. Just do
d(u/v)
then substitute u=C(x),v=x,du=C'(x),dv=dx,x=10

 

[soroban]

Hello, johnjones!

Given that $\displaystyle C(10)\,=\,120$ and $\displaystyle C'(10)\,=\,8,$
find $\displaystyle \frac{d}{dx}\left[\frac{C(x)}{x}\right]$ for $\displaystyle x = 10.$

Do I use the quotient rule? . . . . yes
I don't know the function of C(x). . . . . You don't need it

We have: .$\displaystyle f(x)\:=\:\frac{C(x)}{x}$

Quotient Rule: .$\displaystyle f'(x)\;=\;\frac{x\cdot C'(x)\,-\,1\cdot C(x)}{x^2}$

When $\displaystyle x=10:\;\;f'(10)\;=\;\frac{10\cdot C'(10)\,-\,C(10)}{10^2}$

Since $\displaystyle C(10)=20$ and $\displaystyle C'(10) = 8:\;\;f'(10)\;=\;\frac{10\cdot8\,-\,20}{10^2}\;=\;\frac{60}{100}\;=\;\frac{3}{5}$