Derivative as difference quotient with tricky limits due to square roots I have a function that I want to find the derivative of using the difference quotient definition of a derivative. The function is:
$$f(x)=\frac{\sqrt{x}}{x+1}$$
therefore, using the difference quotient definition, we have:
$$f'(x)=\lim_{h\to0}\frac{1}{h}\left(\frac{\sqrt{x+h}}{x+h+1}-\frac{\sqrt{x}}{x+1}\right)$$
this is equal to:
$$\lim_{h\to0}\frac{1}{h}\left(\frac{(x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}}{(x+1)(x+h+1)}\right)$$
I tried using (a+b)(a-b) = a^2 - b^2 to get rid of the square roots in the numerator, but the denominator gets pretty huge, so I'm not sure if this is a wise path to proceed down.
At this point I get stuck with the algebra. I understand that this is a simple derivative to take using the quotient rule, but I'm trying to practice taking limits, and to learn useful algebra tricks.
A step by step computation would be helpful. Many thanks
 A: Alternatively, you can use the definition
$$f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}.$$
This gives
$$\frac{f(x)-f(a)}{x-a}=\frac{\frac{\sqrt x}{x+1}-\frac{\sqrt a}{a+1}}{x-a}=\frac{(a+1)\sqrt x-(x+1)\sqrt a}{(a+1)(x+1)(x-a)}$$
$$=\frac{(a\sqrt x-x\sqrt a)+(\sqrt x-\sqrt a)}{(a+1)(x+1)(x-a)}=\frac{-\sqrt {ax}\cdot (\sqrt x-\sqrt a)+(\sqrt x-\sqrt a)}{(a+1)(x+1)(x-a)}$$
$$=\frac{1-\sqrt {ax}}{(a+1)(x+1)}\cdot\frac{\sqrt x-\sqrt a}{x-a}$$
As $x \rightarrow a$, the first factor becomes
$$\frac{1-a}{(a+1)^2}$$
and the second
$$\frac{\sqrt x -\sqrt a}{x-a}=\frac{x-a}{(x-a)(\sqrt x +\sqrt a)}=\frac{1}{\sqrt x+\sqrt a}\rightarrow \frac{1}{2\sqrt a}$$
A: You properly wrote$$\lim_{h\to0}\frac{1}{h}\left(\frac{(x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}}{(x+1)(x+h+1)}\right)$$ Now, it is clear that the problem is within $\sqrt{x+h}$. But, you could use the generalized binomial theorem to write $$\sqrt{x+h}=\sqrt{x}+\frac{h}{2 \sqrt{x}}+\cdots$$ So, for the numerator $$A=(x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}=(x+1)\left(\sqrt{x}+\frac{h}{2 \sqrt{x}}+\cdots \right)-(x+h+1)\sqrt{x}$$ Expanding and simplifying $$A=\frac{h}{2 \sqrt{x}}-\frac{h \sqrt{x}}{2}=h\frac{1- x}{2 \sqrt{x}}$$ and the overall problem becomes simple.
A: Using your idea, just ignore the denominator until later. If you go through the numerator, you will see that
$$\begin{align*}
&\left((x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}\right)\cdot\left((x+1)\sqrt{x+h}+(x+h+1)\sqrt{x}\right)\\
=&(x+h)(x+1)^2-x(x+h+1)^2\\
=&hx^2+2hx+h+x^3+2x^2+x-h^2x-2hx^2-2hx-x^3-2x^2-x\\
=&-hx^2-h^2x+h\\
=&-h(x^2+xh-1)
\end{align*}$$
at which point you notice this factor of $h$ will get along nicely with the factor of $\frac{1}{h}$...
The limit then becomes
$$f'(x)=\lim_{h\to 0}-\frac{x^2+xh-1}{(x+1)(x+h+1)\left((x+1)\sqrt{x+h}+(x+h+1)\sqrt{x}\right)}$$
