I have a function f(x) = sqrt(5x+1) and I'm trying to find f'(a) using the definition: lim h->0 f(a+h) - f(a)/h I'm having a hard time simplifying the expression. Help!
Okay. You have: \[\lim_{h \rightarrow 0}\frac{\sqrt{5x+5h+1}-\sqrt{5x+1}}{h}\] Now let me look at it :P
I got it :P I think. Multiply the top and bottom by: \[\sqrt{5x+5x+1}+\sqrt{5x+1}\] After that, see if you can figure it out :P
lim┬(h→0)〖([√(5(a+h)+1)]- [√(5a+1)])/h〗
rationalize!
ratiaonlize the numerator!
multiply by our favorite thingy the conjugate of the top
if we do it to the top we have to do it the bottom
malevolence has it . you have to do some algebra, but the numerator will be \[5x+5h+1-(5x+1)=5h\]
the h will cancel because u have one on bottom too
Then you get: \[\lim_{h \rightarrow 0}\frac{5x+5h+1-5x-1}{h(\sqrt{5x+5h+1}+\sqrt{5x+1})}\] Everything cancels in the top leaving a 5h. Divide out the h in the top and bottom. Then you take the limit and the denominator becomes: \[\sqrt{5x+1}+\sqrt{5x+1}=2\sqrt{5x+1}\] Giving! \[\frac{5}{2\sqrt{5x+1}}\]
and the denominator will be \[h(\sqrt{5x+5h+1}+\sqrt{5x+1})\]
appletastic!
cancel the h wow! nice latex malevolence!
I was trying to type quickly :P Got you and myininaya jumping down my throat xPPP
lol
\[\color{green}{\text{appletini}}\]
omg satellite you will not believe this but malevolence corrected me earlier
latex color race this fall
i tried to tell him it was just a typo like you always say
\[\color{blue}{\text{you probably needed it}}\]
\[\huge\int\limits x^2dx\]
when i do it: typo when you do it: error
wow thats big
LARGE!
the satellite signal is out! come on i will be your sidekick
\[\huge\int\limits \int\limits_D \int\limits \frac{P(z)}{Q(z)}dV\]
\[\huge\text{mine's bigger}\]
\[\huge\color{red} {redder}\]
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