What is the derivative of: y=(1+2^x)/(1-2^x) ??
@satellite73
Quotient rule broski!
\[\Large\rm y'=\frac{\color{royalblue}{(1+2^x)'}(1-2^x)-(1+2^x)\color{royalblue}{(1-2^x)'}}{(1-2^x)^2}\]
pretty sure it's 1 over 4
Where you having trouble? Differentiating the exponentail parts?
I already know how to use it in quotient rule and yes,.. I think exponential part is hard for me
Recall:\[\Large\rm \frac{d}{dx}a^x=a^x(\ln a)\] Example:\[\Large\rm \frac{d}{dx}e^x=e^x(\ln e)\]\[\Large\rm \frac{d}{dx}e^x=e^x(1)\]\[\Large\rm \frac{d}{dx}e^x=e^x\]
you guys are all wrong
So whenever you are taking the derivative of an exponential that's NOT base e, you'll have a log showing up. \[\Large\rm \frac{d}{dx}2^x=2^x(\ln2)\]
But if you want better justification than that, you can look back at logarithmic differentiation.
also, remember to include the point where it is not derivable
Hmm I think this one is differentiable over it's entire domain though, yes? :o Or is that not what you meant? @FibonacciChick666
it shouldn't be at x=0, you have a vertical asympotote
Well that's not in the domain of the function :) But yes, point taken.
true
but it's technically not continuous there ;P
you guys are all wrong
I'm stuck in this part: y'=((1-2^x)(2^xln2)-(1+2^x)(-2^xln2))/(1-2^x)^2
? why are you stuck?
or, what has you stuck?
simplifying XD
Ah you have to distribute and all that jazz >.< Yah it can get a little messy.
oh! well, start with the easy stuff: simplify this (1-2^x)(2^xln2)
how can i multiply \[(-2^xln2)*(2^x)\] ?
what is e^x*e^x?
is it the same as squared?
\[e^(2x)\] yes
so, same here only you have 2^x*2^x
times your minus natural log of 2
no, i mean : (-2^x)*(2^x)
but you don't have that
you have \((-1)(2^x)*(2^x)\)
if you did, that would be as simple as it gets
thanks :D that really help me
np
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