Finding the Derivative of f(a) with Definition

Al3x L3g3nd
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Homework Statement



Find derivative of f(a) for f(t)=(2t+1)/(t+3) using the definition of a derivative

Homework Equations



f '(a)=lim as x goes to a of (f(x)-f(a))/(x-a)

The Attempt at a Solution



f '(a)=lim as x goes to a of (f(x)-f(a))/(x-a)
f '(a)=lim as t goes to a of (((2t+1)/(t+3))-((2a+1)/(a+3)))/(t-a)
f '(a)=lim as t goes to a of (((2t+1)(a+3))/((t+3)(a+3)))-(((2a+1)(t+3))/((a+3)(t+3)))
simplified and got
f '(a)=lim as t goes to a of ((-5a+5t)/((a+3)(t+3)))/(t-a)

I don't know where to go from here.
 
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Al3x L3g3nd said:

Homework Statement



Find derivative of f(a) for f(t)=(2t+1)/(t+3) using the definition of a derivative

Homework Equations



f '(a)=lim as x goes to a of (f(x)-f(a))/(x-a)

The Attempt at a Solution



f '(a)=lim as x goes to a of (f(x)-f(a))/(x-a)
f '(a)=lim as t goes to a of (((2t+1)/(t+3))-((2a+1)/(a+3)))/(t-a)
f '(a)=lim as t goes to a of (((2t+1)(a+3))/((t+3)(a+3)))-(((2a+1)(t+3))/((a+3)(t+3)))
simplified and got
f '(a)=lim as t goes to a of ((-5a+5t)/((a+3)(t+3)))/(t-a)

I don't know where to go from here.

Rewriting your last equation in readable form:$$
\frac{-5a+5t}{(a+3)(t+3)(t-a)}$$You are almost there. Factor a 5 out of the numerator, cancel like factors, and let ##t\to a## and you will have it.
 


LCKurtz said:
Rewriting your last equation in readable form:$$
\frac{-5a+5t}{(a+3)(t+3)(t-a)}$$You are almost there. Factor a 5 out of the numerator, cancel like factors, and let ##t\to a## and you will have it.

wow i feel dumb for not realizing that.

thanks :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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