Question

sorry for the probably simple question, but i haven't done this stuff in a long, long time.
thought i had this right, but again not finding the answer
which of the following gives the slope of the tangent line to the graph of y=2^(1-x) at x=2?
a)-1/2
b)1/2
c)-2
d)2
e)-(ln2)/2

so... it thought y=2^1-x)
then y'=(1-x)(2^(1-x-1))
at x=2
then y'=(1-2)(2^(1-2-1))
y'= -1(2^(-2))
y'= -1/4
any help figuring out where i went wrong would be greatly appreciated.

thank you

Answer 1

y = 2^(1-x) is an exponential function, which is differentiated using natural logarithms

dy/dx = 2^(1-x) (-1) ( ln2)

so when x = 2,
dy/dx = -ln2(2^(1-2)
= -ln2 (1/2) = (-1/2)ln2 , looks like e

as soon as I had my derivative which contained ln2, I knew that the first 4 choices were not correct.

You found the derivative as if it was a polynomial function where the terms are such that the base is the variable, such as 3x^5

e.g.
compare 4^x with x^4

if y = x^4
dy/dx = 4x^3
but
if y = 4^x
dy/dx = ln4 (4^x)

Answer 2

Thanks Reiny,really appreciate the extra explanation. get it now. thank you