What is derivitive of 2^(-x)?

Answer 1

d/dx (a^x) = a^x ln(a)

a = 1/2 since 2^(-x) = [2^(-1)]^x = (1/2)^x

d/dx (2^(-x)) = 2^(-x) * ln(.5)

note that this is the same as -ln(2) * 2^(-x) as indicated by some of the other answers.

Answer 2

This might be the same answer as Scott's. I'm looking at a table of differentials. It says

d(a^u) = a^u ln a du

Plug in a = 2; u = -x; du/dx = -1

2^(-x) ln 2 (-1) = - ln 2 (2^-x) = - (ln 2)/(2^x)

Although 2 = 10/5, so ln 2 = ln 10 - ln 5, don't make that substitution since ln 10 (base e) is not 1.

Answer 3

Think of it like e^(x). Remember that e is just a number (2.7183), Eulers number
Answer
-(2)^(-x) (wrong answer)
Correct answer is -ln(2)*2^(-x)
Can't remember why

Answer 4

-ln(2) 2^(-x)

(negative natural log of 2 times 2 to the negative x)

Explanation:
2^(-x) = (e^(ln(2)))^(-x) = e^[-ln(2)(x)]

Derivative = -ln(2) e^[-ln(2)(x)] = -ln(2) 2^(-x)

Answer 5

-2^(-x) . log(2)