How would I derive this?

Question

anonymous · Answer

Sorry, give me a minute while I type up the equation.

zepdrix · Answer

k c:

anonymous · Answer

$$\lim_{h \rightarrow 0}\frac{ (1+h)^{6}-1 }{ h }$$

zepdrix · Answer

So you certainly could expand out the binomial but that would be messy! You want to recognize that this the limit definition of a function. If we can identify specifically which function that is, we can simply apply the power rule! :D

anonymous · Answer

Yup, this is one the definitions of limits. So I don't have to use the quotient rule?

zepdrix · Answer

Correct c: no quotient rule.

zepdrix · Answer

$$\huge f(x)=x^6$$$$\large f'(x)=\lim_{h \rightarrow 0}\frac{ (x+h)^6-x^6 }{ h }$$

zepdrix · Answer

Hmmmm it looks kind of like this function doesn't it? :o except it's given to us as $$f(1)=1^6$$

zepdrix · Answer

It can be a little tricky to identify what's going on with these limit definitions D: Getting confused on any particular part? <:o

anonymous · Answer

Ah ok, I see what you're saying. So I have to understand that f(1) = 1^6, but would I have to   do algebraic work to arrive at that conclusion? I mean, how would I solve similar problems like this?

zepdrix · Answer

Hmm these are kind of tricky, there really isn't any algebra to go along with it. You just need to recognize WHAT function you're taking the derivative of.

Then skip the whole limit thing and apply the power rule.
In this case, we recognize that the limit is telling us to take the derivative of x^6.

Therefore we can conclude that:
$$\large f'(x)=6x^5$$$$\large f'(1)=6(1)^5=6$$

anonymous · Answer

Ooh ok, I see. Thanks! But how can I apply that strategy to a question like,
$$\lim_{h \rightarrow 0}\frac{ \sqrt[3]{8+h}-2 }{ h }$$
You're right, these limit definition questions are pretty tricky. Thank you for helping me understand them though!

zepdrix · Answer

So the cube root of 8 is what? 2 right? So think of the limit as this...
$$\huge \lim_{h \rightarrow 0}\frac{ (8+h)^{1/3}-8^{1/3} }{ h }$$

zepdrix · Answer

Is it jumping out at you yet? XD hehe

anonymous · Answer

I think I got it:
f(x)=x^(1/3)
f(8)=8^(1/3)
f'(x)=1/(3x^(2/3))
f'(8)=(1/12)

anonymous · Answer

Sorry for not using the special symbols, they can be annoying to use sometimes lol

anonymous · Answer

Yup, I understand it now. I knew I had to work backwards but I didn't know where to start. Thank you so much for your help!

zepdrix · Answer

yay team \c:/

zepdrix · Answer

Your derivative is x^-(2/3) right? :O when you subtracted 1 from it?

zepdrix · Answer

Oh u wrote it as 1 over... nevermind hehe

anonymous · Answer

Yes! And yup!