# Find the derivative of x + sqrtx using first principles, the long way. Help?

Nov 23, 2017

The "first principles" definition of a derivative is:

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

We are given:

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

To obtain $f \left(x + h\right)$, substitute $x + h$ for every x in $f \left(x\right)$:

$f \left(x + h\right) = x + h + \sqrt{x + h}$

Substitute $f \left(x + h\right)$ and $f \left(x\right)$ into the definition:

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

Distribute the negative sign:

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

Combine like terms:

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

Separate into two fractions:

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

The first fraction, $\frac{h}{h}$, becomes 1:

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

We can use the property $\left(a - b\right) \left(a + b\right) = \left({a}^{2} - {b}^{2}\right)$ to eliminate the radicals in the numerator; this means that we must multiply by 1 in the from of $\frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}$:

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

Please observe that we have effectively multiplied by 1 and the radicals in the numerator are eliminated by squaring:

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

The $x - x$ in the numerator becomes 0 :

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

The $\frac{h}{h}$ becomes 1:

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

Now, we may let $h \to 0$:

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

Combine like terms:

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

Nov 24, 2017

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

#### Explanation:

We know:

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

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

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

$f ' \left(x\right) = L i {m}_{h \to 0} \frac{\cancel{\textcolor{red}{x}} + h + \sqrt{x + h} - \cancel{\textcolor{red}{x}} - \sqrt{x}}{h}$

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

We multiply the top and the bottom of the second fraction by the conjugate of the top:

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

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

$f ' \left(x\right) = L i {m}_{h \to 0} \left(1 + \frac{\cancel{\textcolor{red}{x}} + h \cancel{\textcolor{red}{- x}}}{h \left(\sqrt{x + h} + \sqrt{x}\right)}\right)$

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

$f ' \left(x\right) = L i {m}_{h \to 0} \left(1 + \frac{\cancel{\textcolor{red}{h}}}{\cancel{\textcolor{red}{h}} \left(\sqrt{x + h} + \sqrt{x}\right)}\right)$

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

Now, when $h \to 0$ we get:

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

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

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