f(x)=ln(e^x-e^-x), x>0
By applying in turn the Composite and Quotient Rules, find derivative and second derivative.
Thanks.
To find the derivative and second derivative of the given function f(x) = ln(e^x - e^(-x)), we will need to apply the Composite and Quotient Rules.
First, let's take the derivative of f(x) using the Composite Rule. The Composite Rule states that if we have a function g(x) inside another function h(x), then the derivative of h(g(x)) can be found by differentiating the outer function and multiplying it by the derivative of the inner function.
In this case, our outer function is ln(x), and the inner function is e^x - e^(-x).
Using the Chain Rule, the derivative of ln(e^x - e^(-x)) is given by:
f'(x) = (1 / (e^x - e^(-x))) * (d/dx (e^x - e^(-x)))
Now let's differentiate the inner function e^x - e^(-x) using the Quotient Rule. The Quotient Rule states that if we have a function u(x) divided by v(x), then the derivative of u(x)/v(x) can be found by (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2.
For u(x) = e^x - e^(-x), and v(x) = 1, we have:
f'(x) = (1 / (e^x - e^(-x))) * ((e^x - (-e^(-x))) / 1^2)
Simplifying, we get:
f'(x) = 1 / (e^x - e^(-x)) * (2e^x)
Now, let's find the second derivative f''(x) by differentiating the expression we obtained for f'(x).
Using the Quotient Rule again, we have:
f''(x) = ([(e^x - e^(-x)) * 2e^x]' * (e^x - e^(-x)) - 2e^x * [(e^x - e^(-x))]' ) / ((e^x - e^(-x))^2)
Differentiating both components inside the brackets:
f''(x) = ([(2e^x - 2e^(-x)) * 2e^x] * (e^x - e^(-x)) - 2e^x * [(2e^x)] ) / ((e^x - e^(-x))^2)
Simplifying further:
f''(x) = (4e^(2x) - 4) * (e^x - e^(-x)) - 4e^(2x)) / ((e^x - e^(-x))^2)
So, the derivative of f(x) is f'(x) = 1 / (e^x - e^(-x)) * (2e^x), and the second derivative f''(x) is given by f''(x) = ((4e^(2x) - 4) * (e^x - e^(-x)) - 4e^(2x)) / ((e^x - e^(-x))^2).
Please note that the given function f(x) = ln(e^x - e^(-x)) is only valid for x > 0. The derivative and second derivative formulas we obtained are valid within this interval as well.