n/m, got it.

[JohnCU]

This is about derivatives:

why is the derivative of x + c/x = 1 - c/x^2?

When I work it out I get 1 + [(x-c)/(x^2)]...

[[g(x) * f'(x)] - [f(x) * g'(x)]]/(g(x))^2

right?

 

[her209]

Do you know what the derivative of 1/x is ?

 

[JohnCU]

-1/x^2...

 

[her209]

Ummmm okay....

 

[JohnCU]

lol, thanks.

 

[Ionizer86]

1+ (-c)/(x^2) is it, isn't it? Pull the c out then derive x^-1: drop the -1 to the constant, and make the power x^-2.

You have: existing c times the derivative of x^-1: (c) * (1/x^2). Don't need a quotient rule for this, but if you use it:

(f'g - g'f)/g^2 you have: (0)(x) - (1)(c) / (x^2). Because when you derive c you get 0.

<--thinks CalcAB is easy and likes the fig minus gif part of the quotient rule. PS, TI-89 is your friend