Question

If y=1/2( e^x + e^-x ), show that (dy/dx)^2=y^2-1

Answer 1

@Jhannybean

Answer 2

(dy/dx)^2 doesn't mean 2nd derivative there, right?

Answer 3

implicitly differentiate both sides.

Answer 4

So dy/dx by itself means the 1st derivative
So (dy/dx)^2 means the first derivative squared

Answer 5

So you need to find the derivative of y = (1/2)(e^x + e^(-x)) and square it
Then simplify to show that (dy/dx)^2 is the same as y^2 - 1

Answer 6

if you know hyperbolics @eric_d ...

Answer 7

Nope

Answer 8

If you're looking for an easier method than what I have stated above, then you have to compute the derivative either way since you have dy/dx on the left

Answer 9

\[\sinh = \frac{e^x-e^{-x}}{2}\]

Answer 10

why e^x-e^-x / 2 instead of (e^x + e^-x) / 2 ?

Answer 11

You can use hyperbolic trig functions though
y = (1/2)(e^x + e^(-x)) = cosh(x) But you would still need to take the derivative

Answer 12

Oh, whoops.cosine is the positive one, thought it was sine.

Answer 13

you can just as easily take the derivative of each part.

Answer 14

That's what I said ...

Answer 15

he was not introduced to hyperbolics yet.. so may be lets continue with e^x...

Answer 16

\[\frac{d}{dx}(y) = \frac{1}{2} \left[(e^x)' +(e^{-x})'\right]\]