Compute $\frac{d}{dx}(\frac{2x+1}{x+3})$ using definition of derivative.

Question

Problem

Compute $f'(x)$ using definition of derivative when $f$ is defined as:

$$ f(x)=\frac{2x+1}{x+3} $$

Attempt to solve

A derivative is defined as:

$$ f'(a)= \lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h} $$

then we have:

$$ \lim_{h \rightarrow 0} (\frac{2(a+h)+1}{a+h+3}-\frac{2a+1}{a+3})\cdot \frac{1}{h} $$

$$ =\lim_{x \rightarrow 0} \frac{2a+2h+1}{h^2+ah+3h}- \lim_{h \rightarrow 0}\frac{2a+1}{ah+3h} $$

But then i get stuck from here. How do i factor $h^2+ah+3h$ in a way it cancels with $2a+2h+1$. Do they have a factor in common ?

Answer 1

hint

Observe that

$$f(x)=2-\frac{5}{x+3}$$

thus

$$f(a+h)-f(a)=$$ $$5\Bigl(\frac{1}{a+3}-\frac{1}{a+h+3}\Bigr)$$

and

$$\frac{f(a+h)-f(a)}{h}=$$ $$\frac{5}{(a+3)(a+h+3)}$$

The limit when $h\to 0$ is

$$f'(a)=\frac{5}{(a+3)^2}.$$

Answer 2

It's not legal to split a limit into two pieces unless both the pieces exist. In your last displayed expression, you split the limit across a minus sign, but neither of the two resulting limits exist.

Instead, subtract those fractions and multiply out everything in the numerator. Once you cancel everything in the numerator, you'll have

$$\frac{5h}{(a+h+3)(a+h)h}.$$