Calc Question?

Question

Id the deriviative of ( e^ x + e^ -x ) / 2 equal to ( -e^ x - e^ -x ) / 2 and is the second derivative the same as the orig. problem? And what does y' (1) equal?

JV - Hit shift 6 for it to show up

everyone is giving me the same answer, but isn't the derivative of
e^ -x equal to e^-X times -1

Answer 1

The first derivative is:

(e^x - e^-x)/2
Only the second thing becomes negative (because it's e^-x). The first one stays the same because it's e^+x.

To find y'(1), just put in 1 for x. You'll get:

(e^1 - e^-1)/2

which simplifies to:

e/2 - 1/(2e)

And you're right, the second derivative is the same as the original problem!

Answer 2

The derivative of (e^x + e^-x)/2 is equal to (e^x - e^-x)/2 (you got one sign wrong). And yes the second derivative is the same as the original.
y'(1) is the value of the derivative when you have x=1.
So it's equal to (e- (e^-1))/2 which should be equal to about 1,175 (I took e=2,718).

Answer 3

The first derivative is:
Quotient Rule: (2)(e^x-e^-x)-[(e^x+e^-x)(0)]/2^2
2(e^x-e^-x)/4
(e^x-e^-x)/2
Yes, the second derivative is the same as the original problem.
As for y'(1), just substitute (1) wherever you see (x) in the first derivative. y' is another way to write the first derivative.

Answer 4

if y= ( e^ x + e^ -x ) / 2, 1st derivative is:

y' = (e^x - e^-x)/2, so y'(1) = (e^1 - e^-1) / 2 = [e - (1/e)] / 2.

y'' = second derivative = ( e^ x + e^ -x ) / 2 = y, original.

And, y''(1) = y(1).

Answer 5

the first derivative is (e^x-e^(-x))/2 and the second is ( e^ x + e^ -x ) / 2 and Y'(1)=((e^2-1)/2e

Answer 6

let y = (e^ x + e^(-x)) / 2 (= cosh(x))

Then y' = (e^ x - e^(-x)) / 2 (= sinh(x))
and y'' = (e^ x + e^(-x)) / 2 (= cosh(x))

y'(1) = (e^1 - e^(-1)) / 2
= (e² + 1)/(2e)
≈ 1.5431

Answer 7

how do you do that power sign, the one that looks like > but pointing up?