# Using the limit definition, how do you find the derivative of f(x) =sqrt (x+1) ?

Dec 25, 2015

To find the derivative, apply the definition:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

#### Explanation:

With $f \left(x\right) = \sqrt{x + 1}$, we apply the definition:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h}$

To handle the terms in the numerator, it looks like we will need to multiply by the conjugate:

$f ' \left(x\right) =$
${\lim}_{h \to 0} \frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h} \cdot \frac{\sqrt{x + h + 1} + \sqrt{x + 1}}{\sqrt{x + h + 1} + \sqrt{x + 1}}$

f'(x) = lim_(h->0) ((x+h+1) - (x+1))/(h(sqrt(x+h+1)+sqrt(x+1))

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{h}}{\cancel{h} \left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}}$

At this point, we can directly apply the limit to arrive at an answer:

$f ' \left(x\right) = \frac{1}{\sqrt{x + 0 + 1} + \sqrt{x + 1}}$

$f ' \left(x\right) = \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{x + 1}}$

If you are familiar with the chain rule for derivatives, we can use it to test our result:

$f \left(x\right) = \sqrt{x + 1} = {\left(x + 1\right)}^{\frac{1}{2}}$
$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(x + 1\right)}^{\frac{1}{2} - 1}$
$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(x + 1\right)}^{- \frac{1}{2}}$
$f ' \left(x\right) = \frac{1}{2 \sqrt{x + 1}}$