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
$\endgroup$ 34 Answers
$\begingroup$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 $1$.
$\endgroup$ $\begingroup$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
$$ \lim_{h\rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}} $$
$\endgroup$ $\begingroup$$$f'(x)=\frac{dy}{dx}$$ 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'.
Just choose your weapon.
$\endgroup$ $\begingroup$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 $f(x)=\sqrt{x}=x^{\frac{1}{2}}$ $$f'(x)=\frac{dy}{dx}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2\sqrt{x}}$$
$\endgroup$ 4
