How do you differentiate 2^x? Thanks!?

Question

I'm thinking its ln2 but not sure...

Answer 1

y = 2^x

ln y = ln 2^x

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

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

Pull the constant (ln 2) out:

y'/y = ln 2 * d/dx x

y'/y = ln 2

y' = y * ln 2

y' = 2^x * ln 2

*** Remember this advice, kids. Whenever you want to get rid of a variable exponent during differentiation, use logarithmic differentiation. Using it will make things cleaner and easier.

Answer 2

How can everyone get this wrong?

d/dx (2^x) = d/dx (e^(x ln 2)) = e^(x ln 2) ln 2 = 2^x ln 2

Errors in the other answers:

rubberbandbanker - your method is valid, but in your last step:
"y'/y=ln 2
y'=(xln2)*(ln2)"

y=2^x, not x ln 2 (x ln 2 is ln y, not y)

jj - power rule applies when the base is a variable and the power is a constant, not the other way around

izzyhearts - your formula is transposed. It's a^x ln a, not a ln a^x

Answer 3

Surely its x*2^(x-1). It can't be ln2 as ln2=0.693.... which is a constant, and presumably x is a variable?

Answer 4

u take the ln of both sides, then differentiate
lny=xln2
then differentiate
y'/y=ln2
so then u plug in y for y and multiply it so u isolate y'
y'=(x^2)*(ln2)
its right now sorry i made a careless mistake before

Answer 5

y = x^(2x - 3) We are going to find dy using logarithmic differentiation, so first take the natural log of each side: ln(y) = (2x - 3)ln(x) Take the derivative of each side, the left by the log properties for derivatives and the right by the product rule: dy/y = (2x - 3) * 1/x + 2ln(x), which can also be written: dy/y = 2 - 3/x + 2ln(x) Now, multiply both sides by y: dy = y * (2 - 3/x + 2ln(x)) Plug back in, and we're done: dy = (x^(2x - 3)) * (2 - 3/x + 2ln(x))

Answer 6

y=2^x
lny=ln(2^x)
lny=xln2
(dlny/dy)*dy/dx = ln2 dx/dx
(1/y)*dy/dx=ln2
dy/dx=y*ln2
dy/dx=(2^x)*ln2
hope this helps
ur answer is not right
here what u r doing is applying implicit differentiation and the chain rule.

Answer 7

the answer is 2 ln(2^x). i think.
the formula is a ln(a^x), where a is the coefficient

Answer 8

y=2^x

log y = log(2^x)
logy = x* log2
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