Thread: Derivative of another function's derivative?

So we just started getting into derivatives in class, and I understand the concept and the formulas. I've read through my textbook several times and can't come across a similar problem. I'm sure I'm just overlooking something I know. But the more I stare at this question the angrier I get because it seems so simple. Also, if you could tell me what this 'type' of problem is, I'd appreciate it.

Consider the function h(x), for which h(5) = 2 and h'(5) = 4.
Let f(x) = h(x) / x. Then f'(5) = ?

Thanks in advance.

(using the quotient rule)f'(x) = $\displaystyle \frac{ h'(x)x - h(x)}{x^2} $ so f'(5) = $\displaystyle \frac{ h'(5)5 - h(5)}{5^2} = \frac{4*5 - 2}{25} $ so f'(5) = $\displaystyle \frac{18}{25} $

Thanks I was trying to set up something like that but I was having issues because the formula given in class for quotient functions was as follows:
$\displaystyle \frac{numderiv * denom - num * denom deriv}{denom^2} $ and I was getting hung up because the derivative of x wasn't defined. I forgot that it was always 1. Is that a correct understanding? Thanks again so much. You've made my night!

Originally Posted by jjtjp

Thanks I was trying to set up something like that but I was having issues because the formula given in class for quotient functions was as follows:
$\displaystyle \frac{numderiv * denom - num * denom deriv}{denom^2} $ and I was getting hung up because the derivative of x wasn't defined. I forgot that it was always 1. Is that a correct understanding? Thanks again so much. You've made my night!

yes

In fact, the "derivative" is a generalization of the "slope" of a straight line so the derivative of any linear function, ax+ b is just the constant slope, a. Here, a= 1 and b= 0.