derivative of 2^x ?

Question

What is the derivative of 2^x ? Please explain

Answer 1

f(x) = a^x, a = constant

f'(x) = ln a * a^x

d/dx 2^x = ln 2 * 2^x

Answer 2

I would think that the derivative of 2^x would be x*2^(x-1)

Answer 3

Remember that you can write a number, say "c", as e^(ln(c)). So 2^x = e^(ln(2^x)) = e^(ln(2) x). You can easily take the derivative of this.

Answer 4

Look at the beautifully formatted and easy to read PDF solution at the link below.

http://www.tomsmath.com/derivative-of-2x.html

Answer 5

Another way: 2^x=e(log (2^x).
Using properties of logs =e^(x*log 2)
Taking the derivative with the chain rule (d(e^y)/dx=dy/dx*e^y gives log 2*e^(x*log 2) as d(x*c)=c

Simplifying: log 2 * e^log(2^x) or log 2*2^x

Answer 6

Use logarithmic differentiation

ln y = x ln2

y'/y = ln2

y' = yln2

y' = 2^x * ln(2)

Answer 7

let 2^x = y

taking logs

x log 2 = log y

diffrentiating

log 2 dx = (1/y) dy

dy = y(log2)dx

dy/dx = y log2

but y = 2^x

dy/dx = (log2)2^x

Answer 8

2^x ln(2)

Answer 9

y=2^x
lny=xln2
y'/y=ln2
y'=yln2=2^xln2

Answer 10

y = 2^x
log y = x log 2
(1/y) (dy/dx) = log 2
dy/dx = y log 2
dy/dx = (2^x) log 2
Let log be log to base 2
dy/dx = 2^x