Calculate this derivative by definition: [tex] \frac{1}{

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nathangirnet

04.08.2019

Let $f(x) = \dfrac{1}{\sqrt{x}}$ so ze definition of the derivative is $\displaystyle f'(x) = \lim_{h\to0} \dfrac{f(x+h) - f(x)}{h}$ since $f(x) = \tfrac{1}{\sqrt{x}}$ and $f(x+h) = \frac{1}{\sqrt{x+h}}$ $\displaystyle f'(x) = \lim_{h\to0} \dfrac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}$ we can't shove in $h=0$ yet because we will divide by 0. let us multiply the numerator by its conjugate to rationalize those denominators (no radicals)the conjugate of $\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}$ is $\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}$ . let us multiply the numerator and denominator by that now so that we can rationalize and see what happens. $\begin{aligned} f'(x) & = \lim_{h\to0}\left[ \dfrac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \cdot \dfrac{\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}}\right] \\ & = \lim_{h\to0}\left[ \dfrac{\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right)\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)}{h\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)} \right] \end{aligned}$ currently we have too many things to deal with. let us focus on the numerator. $\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right)\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)$ since difference of squares factoring is $(a+b)(a-b) = a^2 - b^2$ , multiplying the two conjugates gets us $\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right)\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right) = \left(\frac{1}{\sqrt{x+h}}\right)^2 - \left(\frac{1}{\sqrt{x}}\right)^2$ which simplifies to $\frac{1^2}{\left(\sqrt{x+h}\right)^2} - \frac{1^2}{\left(\sqrt{x}\right)^2}= \frac{1}{x+h} - \frac{1}{x}$ if we want to combine these fractions, we need a common denominator.multiply numerator and denominator of $\frac{1}{x+h}$ by $x$ multiply numerator and denominator of $\frac{1}{x}$ by $(x+h)$ $\begin{aligned} \frac{1}{x+h} - \frac{1}{x} & = \frac{1 \cdot [x]}{(x+h) \cdot [x]} - \frac{1 \cdot [x+h] }{x \cdot [x+h] } \\ & = \frac{x}{x(x+h)} - \frac{(x+h) }{x (x+h) } \\ & = \frac{x - (x+h)}{x(x+h)} \\ & = \frac{x - x-h}{x(x+h)} \\ & = \frac{-h}{x(x+h)} \end{aligned}$ therefore, we plug this back into our difference quotient. then we rewrite the division as multiplication. $\begin{aligned} f'(x) & = \lim_{h\to0} \left[ \dfrac{\frac{-h}{x(x+h)}}{h\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)} \right] \\ & =\lim_{h\to0} \left[ \frac{-h}{x(x+h)} \cdot \dfrac{1}{h\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)}\right] \end{aligned}$ we can cancel out the h. $\displaystyle f'(x) = \lim_{h\to0} \left[ \dfrac{-1}{x(x+h)} \cdot \dfrac{1}{\left(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\right)}\right]$ we are now able to take the limit by doing direct substitution. $\begin{aligned} f'(x) & =\left[ \dfrac{-1}{x(x+0)} \cdot \dfrac{1}{\left(\frac{1}{\sqrt{x+0}} + \frac{1}{\sqrt{x}}\right)}\right] \\ & = \dfrac{-1}{x^2} \cdot \frac{1}{ \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}}} \\ & = \dfrac{-1}{x^2} \cdot \frac{1}{ \frac{2}{\sqrt{x}}} \\ & = \dfrac{-1}{2\cdot \frac{x^2}{\sqrt{x}}} \end{aligned}$ we can rewrite $\sqrt{x} = x^{1/2}$ $\begin{aligned} f'(x) & = \dfrac{-1}{2\cdot \frac{x^2}{x^{1/2}} } \\ & = \dfrac{-1}{2\cdot x^{2-\frac{1}{2}}} \\ & = \dfrac{-1}{2x^{3/2}} \end{aligned}$ thus $\dfrac{d}{dx} \left[\frac{1}{\sqrt{x}}\right] = \dfrac{-1}{2x^{3/2}}$

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Calculate this derivative by definition: $\frac{1}{ \sqrt{x} }$

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