Question

d/dx 2^x sinx = ?

Answer 1

I believe that the ans. is ....2^x(Ln2sinx + cosx) ? how would I the solve for X??

Answer 2

d/dx 2^x sinx = 2^x (ln 2 sin x +cos x)
yes!
but why do u need to solve for x ?

Answer 3

To find the critical points for the minima and maxima

Answer 4

2^x can't be 0
so,
(ln 2 sin x +cos x) =0
ln 2 sin x = -cos x
cot x = -ln 2

Answer 5

so, a critical point is

cot inverse (ln 1/2)

Answer 6

The answers given in the book are X= -0.96,2.18,5.32 I am supposed to find the CP's between theinterval of {-2,6}

Answer 7

yes! yes
cot inverse (ln 1/2) = -0.96
if you use calculator

and since cot function is periodic in pi ,
other solutions are
-0.96 +pi =2.18
-0.96+2pi =5.32
:)

Answer 8

I can see how you set things up but I guess that I am not that good with the calc.

I am using a TI 84 so if I wanted cot inv how would I enter that?

Answer 9

i have no idea about TI 84.....sorry
know how to do tan inverse in that ?

Answer 10

because cot inverse x = tan inverse 1/x

Answer 11

i did it like this, here
http://www.wolframalpha.com/input/?i=arccot+ln+0.5

Answer 12

good ol WR I got .96 using Tan inverse 1/ln1 I do not know why I did not get a negative but I will work on it. thx

Answer 13

welcome ^_^

Answer 14

thanks again!