Grace is planning to install grass in the yard shown above. She will not be putting the grass in the two circular flower beds or the rectangular pool.

Grace is planning to install grass in the yard shown above. She will not be putting the grass in th

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nathangirnet

08.03.2022

Mathematics

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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