# How do you find the instantaneous rate of change of 2/(sqrt(x) + 1) using the lim_(h->0) method?

Aug 18, 2017

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

See explanation for method.

#### Explanation:

The limit definition of the derivative states that the derivative of a function $f$ is given by:

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

So here, where $f \left(x\right) = \frac{2}{\sqrt{x} + 1}$, we see that:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\frac{2}{\sqrt{x + h} + 1} - \frac{2}{\sqrt{x} + 1}}{h}$

Now, just power through some algebra:

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{\frac{\sqrt{x} + 1 - \left(\sqrt{x + h} + 1\right)}{\left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right)}}{h}$

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right)}$

Multiply the numerator and denominator by the conjugate of the numerator:

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{\left(\sqrt{x} - \sqrt{x + h}\right) \left(\sqrt{x} + \sqrt{x + h}\right)}{h \left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right) \left(\sqrt{x} + \sqrt{x + h}\right)}$

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{x - \left(x + h\right)}{h \left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right) \left(\sqrt{x} + \sqrt{x + h}\right)}$

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{- h}{h \left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right) \left(\sqrt{x} + \sqrt{x + h}\right)}$

$f ' \left(x\right) = 2 {\lim}_{h \rightarrow 0} \frac{- 1}{\left(\sqrt{x + h} + 1\right) \left(\sqrt{x} + 1\right) \left(\sqrt{x} + \sqrt{x + h}\right)}$

Now we can evaluate the limit, since $h$ is out of the denominator:

f'(x)=(-2)/((sqrt(x+0)+1)(sqrtx+1)(sqrtx+sqrt(x+0))

f'(x)=(-2)/((sqrtx+1)(sqrtx+1)(sqrtx+sqrtx)

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