Posted by **ken** on Saturday, March 2, 2013 at 10:53pm.

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

- calculus -
**Reiny**, Saturday, March 2, 2013 at 11:51pm
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)

- calculus -
**ken**, Sunday, March 3, 2013 at 12:37am
Thanks Reiny,really appreciate the extra explanation. get it now. thank you

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