How do I prove $\frac{ \sqrt{x+h}-\sqrt{x} }{ h}=\frac{1}{\sqrt{x+h}+\sqrt{x}}$?

Question

$$\frac{ \sqrt{x+h}-\sqrt{x} }{ h}=\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

I know I just asked a question and I did figure out how that one worked but I'm not sure how I would go about this one.

Answer 1

\begin{align} \frac{1}{\sqrt{x+h}+\sqrt{x}} &= \frac{1}{\sqrt{x+h}+\sqrt{x}} \cdot \frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}-\sqrt{x}} \\ &= \frac{\sqrt{x+h}-\sqrt{x}}{(x+h)-x} \\ &= \frac{\sqrt{x+h}-\sqrt{x}}{h} \end{align}

Answer 2

Hint: Multiply by $$ 1 = \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} $$

Answer 3

Multiply the top and bottom by $\sqrt{x+h} + \sqrt{x}$.

Answer 4

Observe that $$\begin{align}\frac{\sqrt{x+h}-\sqrt{x}}{h}&=\frac{\sqrt{x+h}-\sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ &=\frac{1}{\sqrt{x+h}+\sqrt{x}}\end{align}$$ by multiplying by $1$.

Answer 5

Just like DanZimm said, multiply by numerator conjugate:

$$\frac{\sqrt{x+h}-\sqrt x}h \cdot \frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x} = \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt x)} = \frac 1{\sqrt{x+h}+\sqrt x}$$