Calcula la derivada de la funcion f(x)=2^x9

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$\left(x\right)=2^x\cdot\:9$(x)=2x· 9

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$\mathrm{Prepare}\:\left(x\right)=2^x9\:\mathrm{for\:Lambert\:form}:\quad xe^{-\ln\left(2\right)x}=9$Prepare (x)=2x9 for Lambert form:    xe−ln(2)x=9

$\mathrm{Rewrite\:the\:equation\:with\:}-x\ln\left(2\right)=u\mathrm{\:and\:}x=-\frac{u}{\ln\left(2\right)}$Rewrite the equation with −xln(2)=u and x=−

u

ln(2)

$\left(-\frac{u}{\ln\left(2\right)}\right)e^u=9$(−

u

ln(2) )eu=9

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$\mathrm{Rewrite}\:\left(-\frac{u}{\ln\left(2\right)}\right)e^u=9\:\mathrm{in\:Lambert\:form}:\quad e^uu=-9\ln\left(2\right)$Rewrite (−

u

ln(2) )eu=9 in Lambert form:    euu=−9ln(2)

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$\mathrm{Solve\:}\:e^uu=-9\ln\left(2\right):\quad\mathrm{No\:Solution\:for}\:u\in\mathbb{R}$Solve  euu=−9ln(2):    No Solution for u∈ℝ

$\mathrm{No\:Solution\:for}\:x\in\mathbb{R}$

Step-by-step explanation:

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

function 2 has a greater rate of change than function 1

function 2 has a greater y-intercept than function 1

step-by-step explanation: