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?
so... it thought y=2^1-x)
any help figuring out where i went wrong would be greatly appreciated.
- 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
compare 4^x with x^4
if y = x^4
dy/dx = 4x^3
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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