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Mathematics 9 Online
OpenStudy (anonymous):

What is the derivative of: y=(1+2^x)/(1-2^x) ??

OpenStudy (anonymous):

@satellite73

zepdrix (zepdrix):

Quotient rule broski!

zepdrix (zepdrix):

\[\Large\rm y'=\frac{\color{royalblue}{(1+2^x)'}(1-2^x)-(1+2^x)\color{royalblue}{(1-2^x)'}}{(1-2^x)^2}\]

OpenStudy (anonymous):

pretty sure it's 1 over 4

zepdrix (zepdrix):

Where you having trouble? Differentiating the exponentail parts?

OpenStudy (anonymous):

I already know how to use it in quotient rule and yes,.. I think exponential part is hard for me

zepdrix (zepdrix):

Recall:\[\Large\rm \frac{d}{dx}a^x=a^x(\ln a)\] Example:\[\Large\rm \frac{d}{dx}e^x=e^x(\ln e)\]\[\Large\rm \frac{d}{dx}e^x=e^x(1)\]\[\Large\rm \frac{d}{dx}e^x=e^x\]

OpenStudy (anonymous):

you guys are all wrong

zepdrix (zepdrix):

So whenever you are taking the derivative of an exponential that's NOT base e, you'll have a log showing up. \[\Large\rm \frac{d}{dx}2^x=2^x(\ln2)\]

zepdrix (zepdrix):

But if you want better justification than that, you can look back at logarithmic differentiation.

OpenStudy (fibonaccichick666):

also, remember to include the point where it is not derivable

zepdrix (zepdrix):

Hmm I think this one is differentiable over it's entire domain though, yes? :o Or is that not what you meant? @FibonacciChick666

OpenStudy (fibonaccichick666):

it shouldn't be at x=0, you have a vertical asympotote

zepdrix (zepdrix):

Well that's not in the domain of the function :) But yes, point taken.

OpenStudy (fibonaccichick666):

true

OpenStudy (fibonaccichick666):

but it's technically not continuous there ;P

OpenStudy (anonymous):

you guys are all wrong

OpenStudy (anonymous):

I'm stuck in this part: y'=((1-2^x)(2^xln2)-(1+2^x)(-2^xln2))/(1-2^x)^2

OpenStudy (fibonaccichick666):

? why are you stuck?

OpenStudy (fibonaccichick666):

or, what has you stuck?

OpenStudy (anonymous):

simplifying XD

zepdrix (zepdrix):

Ah you have to distribute and all that jazz >.< Yah it can get a little messy.

OpenStudy (fibonaccichick666):

oh! well, start with the easy stuff: simplify this (1-2^x)(2^xln2)

OpenStudy (anonymous):

how can i multiply \[(-2^xln2)*(2^x)\] ?

OpenStudy (fibonaccichick666):

what is e^x*e^x?

OpenStudy (fibonaccichick666):

is it the same as squared?

OpenStudy (anonymous):

\[e^(2x)\] yes

OpenStudy (fibonaccichick666):

so, same here only you have 2^x*2^x

OpenStudy (fibonaccichick666):

times your minus natural log of 2

OpenStudy (anonymous):

no, i mean : (-2^x)*(2^x)

OpenStudy (fibonaccichick666):

but you don't have that

OpenStudy (fibonaccichick666):

you have \((-1)(2^x)*(2^x)\)

OpenStudy (fibonaccichick666):

if you did, that would be as simple as it gets

OpenStudy (anonymous):

thanks :D that really help me

OpenStudy (fibonaccichick666):

np

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