# Question 6f00a

##### 1 Answer
Jul 31, 2015

${y}^{'} = - \frac{1}{2} \cdot {x}^{- \frac{3}{2}}$

#### Explanation:

In order to differentiate this function from the first principle, you need to use the formula

color(blue)(d/dx(y) = lim_(h->0)(f(x+h) - f(x))/h

In your case, you will have

$\frac{d}{\mathrm{dx}} \left(y\right) = {\lim}_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}$

$\frac{d}{\mathrm{dx}} \left(y\right) = {\lim}_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \cdot \sqrt{x} \cdot \sqrt{x + h}}$

Multiply this expression by

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

to get

d/dx(y) = lim_(h->0)[(sqrt(x) - sqrt(x+h))(sqrt(x) + sqrt(x+h))]/(h * sqrt(x) * sqrt(x+h) * (sqrt(x) + sqrt(x+h))#

Use the fact that

$\textcolor{b l u e}{\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}}$

to simplify this expression

$\frac{d}{\mathrm{dx}} \left(y\right) = {\lim}_{h \to 0} \frac{{\left(\sqrt{x}\right)}^{2} - {\left(\sqrt{x + h}\right)}^{2}}{\left(h \cdot \sqrt{x} \cdot \sqrt{\cdot} x\right) \cdot \sqrt{x + h} + h \cdot \sqrt{x} \cdot \sqrt{x + h} \cdot \sqrt{x + h}}$

$\frac{d}{\mathrm{dx}} \left(y\right) = {\lim}_{h \to 0} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - h}{h \cdot x \cdot \sqrt{x + h} + h \cdot \left(x + h\right) \cdot \sqrt{x}}$

$\frac{d}{\mathrm{dx}} \left(y\right) = {\lim}_{h \to 0} \frac{- \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{h}}}}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{h}}} \cdot x \cdot \sqrt{x + h} + \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{h}}} \cdot \left(x + h\right) \cdot \sqrt{x}}$

$\frac{d}{\mathrm{dx}} \left(y\right) = - \frac{1}{x \cdot \sqrt{x + 0} + \left(x + 0\right) \cdot \sqrt{x}}$

$\frac{d}{\mathrm{dx}} \left(y\right) = - \frac{1}{2 x \sqrt{x}} = - \frac{1}{2} \cdot \frac{1}{\sqrt{{x}^{3}}} = \textcolor{g r e e n}{- \frac{1}{2} \cdot {x}^{- \frac{3}{2}}}$