Derivative of the Square Root of X [closed]

Given $y = \sqrt x$ and nothing more, using the formula of a limit $$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$ (that is, f prime of x equals the limit of h approaching zero with the equation ((f of the sum of x and h) minus (function of x)) over h)

how do we convert (not evaluate) it into Leibniz's notation, $\frac{dy}{dx}$?

I'm having a lot of problems here, and I hate taking radicals out of denominators. Would someone walk me through this?

(Also, what is MathJaX?)

Question 22 on page 107 of the book from this problem here

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• What do you mean "convert it into Leibniz's notation"? $\frac {df}{dx} = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ by definition. – user269351 Sep 11 '15 at 23:42
• possible duplicate of Trying to show derivative of $y=x^\frac{1}{2}$ using limit theorem – Dylan Sep 11 '15 at 23:50
• @Nefer007 I see that you've edited your question. Again, "convert it into Leibniz's notation" doesn't mean anything: $\dfrac{df}{dx}$ is defined to mean the limit you put in your first line. So if you're not looking for help evaluating that limit for this particular function, then you need to make it clear exactly what you are asking. To your second question, MathJax is the TeX based typesetting library that we use here. See this for more info. – user269351 Sep 12 '15 at 1:12

Here's how to evaluate the limit, in case that's what you're asking: \begin{align}\lim_{h\to 0} \dfrac {f(x+h)-f(x)}{h} &= \lim_{h\to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\ &= \lim_{h\to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h}\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &= \lim_{h\to 0} \dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h\to 0} \dfrac{1}{(\sqrt{x+h}+\sqrt{x})} \\ &= \dfrac{1}{2\sqrt{x}}\end{align}

It's useful to keep in mind that two of the greatest mathematical tricks are multiplication by $1$ and addition by $0$. Above I just multiplied by a particularly useful version of the number $1$.

We want to evaluate

$$\lim_{h\rightarrow 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$

Multiply the top and bottom by $\sqrt{x+h} + \sqrt{x}$ to find that this is equal to

$$\lim_{h\rightarrow 0} \frac{x+h-x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h\rightarrow 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$

Cancelling the $h$ from top and bottom, this is

$$\lim_{h\rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}}$$

$$f'(x)=\frac{dy}{dx}$$ for $$y=f(x)$$ They are two different ways to write the same thing. No conversion necessary, any more than one would need to convert '$5$' to Roman numeral 'V'.

Why don't we find the general form of the power rule and then solve your problem? let $$f(x)=x^n$$ where $n$ is any real number, let's find $\frac{dy}{dx}$ or similarly $f'$ using The Equation: $$f'(x)=\frac{dy}{dx}= \lim \limits_{h \to 0} \left( \frac{f(x+h)-f(x)}{h} \right)$$ $$=\lim \limits_{h \to 0} \left( \frac{(x+h)^n - x^n}{h} \right)$$ $$=\lim \limits_{h \to 0} \left( \frac{\sum_{r=0}^{n} \left( {{n}\choose{r}}x^{n-r}h^r \right) - x^n }{h} \right)$$ $$=\lim \limits_{h \to 0} \left( \frac{(x^n+nx^{n-1}h...+h^n) - x^n}{h} \right)$$ $$=\lim \limits_{h \to 0} \left( \frac{nx^{n-1}h...+h^n}{h} \right)$$ $$=\lim \limits_{h \to 0} \left( nx^{n-1} \right) = nx^{n-1}$$ Now let $f(x)=\sqrt{x}=x^{\frac{1}{2}}$ $$f'(x)=\frac{dy}{dx}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2\sqrt{x}}$$
• I think there is an error in assuming ${n\choose r}$ is defined for fractional $n$... This proof only holds for $n\in\mathbb{N}$ – nathan.j.mcdougall Sep 12 '15 at 5:33
• @ChaosSXDemon zweigmedia.com/RealWorld/proofs/powerruleproof.html outlines the general way to go about extending the rule to the reals (actually it works for complex variables too), running with the idea that $x^n=e^{n\ln(x)}$. This of course appeals to your specific definition of $e$, for which math.stackexchange.com/questions/199447/… would provide some comfort. It's a good question. – nathan.j.mcdougall Sep 12 '15 at 6:51