Help with f ' (a) and tangent lines

[grollio]

I'm taking calculus online and I need help figuring out how to solve the following problem. My book and notes given from my teacher do not show how to solve problems like this and I keep ending up with answers that don't work. The question is;

Find f '(a) when f(t)=(3t+3)/(t+6)







I know I need to use the equation f '(a)= lim as h--> 0 of f(a+h)-f(a)/h




When I attempted this problem I kept getting zero in the denominator which i know is not correct.

 

[Bohrok]

Once you have an expression with the difference quotient, you multiply things out and factor out an h in the numerator which cancels with the h in the denominator. Then you can let h = 0
Also, make sure that (f(a+h)-f(a)) is all in the numerator.

 

[Office_Shredder]

Do you have anything like the quotient rule to use to simplify this problem? It's fairly rare you want to calculate a derivative via the definition

 

[queenofbabes]

Yeah, in practice you almost never calculate a derivative from first principles...Use product/quotient rule (you know what that is?)

 

[HallsofIvy]

Yes, the derivative of f(x), at x= a, it defined as
$$\lim_{h\rightarrow h}\frac{f(a+h)- f(a)}{h}$$
and the denominator always goes to 0! But evaluating the limit is NOT just a matter of putting h= 0. Contrary to popular belief among students, the limit is NOT just a fancy way of talking about the value of the function!

If
$$f(t)= \frac{3t+3}{t+6}$$
then
$$f(a)= \frac{3a+3}{a+ 6}$$
and
$$f(a+h)= \frac{3(a+h)+3}{a+ h+ 6}= \frac{3a+ 3h+ 3}{a+ h+ 6}$$
so
$$f(a+h)- f(a)= \frac{3a+ 3h+ 3}{a+ h+ 6}- \frac{3a+3}{a+ 6}$$
To subtract those fractions, we need to get a common denominator by multiplying the numerator of the first fraction by a+ 6 and of the second fraction by a+h+ 6:
$$f(a+h)- f(a)= \frac{(3a+3h+ 3)(a+6)}{(a+h+6)(a+6)}-\frac{(3a+3)(a+h+6)}{(a+h+6)(a+6)}$$
$$= \frac{((3a^2+3ah+3a+18+ 18h+ 18)- (3a^2+3a+ 3ah+3h+ 18a+ 18)}{(a+h+6)(a+6)}= \frac{18h}{(a+h+6)(a+ 6)}$$
so that when we divide by h we have an h in both numerator and denominator:
$$\frac{f(a+h)-f(a)}{h}= \frac{18h}{(a+h+6)(a+ 6)h}$$
and, as long as h is not 0, we can cancel the h s.
$$\frac{f(a+h)-f(a)}{h}= \frac{18}{(a+h+6)(a+ 6)}$$
and now we can let h= 0 without any problem:
the limit is $18/(a+6)^2$.


I once taught a course called "Calculous for Business Administration and Economics" using a text that had been selected by the Business administartion department. The text had, on one page the "laws of limits":
1) $\lim_{x\rightarrow a} (f(x)+ g(x))= \lim_{x\rightarrow a}f(x)+ lim_{x\rightarrow a}g(x)$

2) [itex]lim_{x\rightarrow a} f(x)g(x)= (lim_{x\rightarrow a}f(x))(lim_{x\rightarrow a}g(x))[/tex]<br /> <br /> 3) $$lim_{x\rightarrow a}\frac{f(x)}{g(x)}= \frac{lim_{x\rightarrow a}f(x)}{lim_{x\rightarrow a} g(x)}$$<br /> <b>as long as $lim_{x\rightarrow a} g(x)$ is not 0</b>.<br /> <br /> On the <b>next</b> page they define the derivative, completely ignoring the fact that, in that definition, the denominator <b>always</b> goes to 0 so those laws don't apply!<br /> <br /> It is essential to have also this very important but often overlooked law: If, in some neighborhood of a, f(x)= g(x) for all x <b>except</b> possibly x=a, then $\lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} g(x)$. That makes it possible to cancel the h's and then take the limit.[/itex]