Find the derivative of

y = (e^x-e^-x)/2

(1/2)(e^x + e^x)

= e^x

I think Damon missed that negative sign

This is a catenary and its derivative is

y' = (1/2) (e^x - e^-x)

To find the derivative of y = (e^x - e^-x)/2, we can use the quotient rule.

The quotient rule states that for a function in the form of u/v, where u and v are both functions of x, the derivative is given by (v * u' - u * v') / v^2.

In this case, we have u = e^x - e^-x and v = 2.

First, let's find u':
To find the derivative of u = e^x - e^-x, we can use the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then the derivative is given by dy/dx = f'(g(x)) * g'(x).

In this case, f(x) = e^x - e^-x and g(x) = x.

The derivative of f(x) = e^x - e^-x with respect to x can be found by applying the chain rule:
f'(x) = (d/dx)(e^x) -(d/dx)(e^-x)
= e^x - (-e^-x)
= e^x + e^-x

Since g(x) = x, g'(x) = 1.

Therefore, u' = (e^x + e^-x) * 1 = e^x + e^-x.

Next, let's find v':
Since v = 2, v' = 0 (the derivative of a constant is always zero).

Now, we can find the derivative of y using the quotient rule:
y' = (v * u' - u * v') / v^2
= (2 * (e^x + e^-x) - (e^x - e^-x) * 0) / 2^2
= (2e^x + 2e^-x) / 4
= (e^x + e^-x) / 2.

Thus, the derivative of y = (e^x - e^-x)/2 is y' = (e^x + e^-x) / 2.