[eqn]\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/eqn]
>>16803116it do be that way
>>16803125uhm actually sweaty it don't be that be that way in lebesque world
>>16803116>doesn't define what f is>doesn't define what y is>doesn't say from what space we take x and h, and what metric we are usingYes, OP, you are a very smart cookie indeed
>>16803116Plug in x=y=2.[eqn]1 = \frac{d2}{d2} = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = 0[/eqn]
>>16803804
>>16803697thisOP has a normal brain, not a giga brain
>>16803116[eqn]f'(x) = \left.\frac{d}{dh}f(x+h)\right|_{h=0}[/eqn]change my mind
>>16803116[eqn]\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/eqn]
>>16803116>He still uses one sided slope approximations[eqn]\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x-h)}{h}[/eqn]
>>16804508Don't you mean[eqn]\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x-h)}{2h}[/eqn]
[math]f(x)=x^n \\\displaystyle\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{(x+h)^n -x^n}{h} \\\displaystyle= \lim_{h \to 0} \dfrac{ \displaystyle \sum_{k=0}^{n} \binom{n}{k}x^{n-k}h^{k} - x^n}{h} = \lim_{h \to 0} \dfrac{ \displaystyle \binom{n}{0}x^n + \binom{n}{1}x^{n-1}h + \sum_{k=2}^{n} \binom{n}{k}x^{n-k}h^{k} - x^n }{h} \\\displaystyle= \lim_{h \to 0} \dfrac{ \displaystyle n x^{n-1}h + \sum_{k=2}^{n}\binom{n}{k}x^{n-k}h^{k}}{h}= n x^{n-1}[/math]
>>16803116I live in bizarro world where [math]\frac{dy}{dx}=\lim_{h\rightarrow \infty}\frac{h}{f(x-h)+f(x)}[/math]
>>16803804Let 2 = II = y and 1 = I = x[math] \frac{dII}{dI} = \frac{dI^2}{dI} = 2 = \lim_{h \to 0} \frac{f(I + h) - f(I)}{h} = 0[/math]
>>16804315uhm no
>>16805835what on fuck
