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 of the given function, we will use the composite and quotient rules.
First, let's simplify the function f(x) = ln(e^x - e^(-x)):
f(x) = ln(e^x - 1 / e^x)
Now, we will find the derivative using the composite rule.
Let u(x) = e^x - 1, and v(x) = 1 / e^x.
Applying the composite rule, we have:
f'(x) = (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2
Now, let's find the first derivative.
u(x) = e^x - 1
u'(x) = d/dx(e^x) - d/dx(1)
= e^x
v(x) = 1 / e^x
v'(x) = d/dx(1 / e^x)
= -1 / (e^x)^2 * (-e^x)
= e^x / (e^x)^2
= e^x / e^(2x)
= 1 / e^x
Now, we can substitute these values into the derivative formula:
f'(x) = (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2
= [(1 / e^x) * e^x - (e^x - 1) * (1 / e^x)] / [(1 / e^x)]^2
= [1 - e^x + 1] / (1 / e^x)^2
= (2 - e^x) * e^2x
The first derivative of the function f(x) = ln(e^x - 1 / e^x) is f'(x) = (2 - e^x) * e^2x.
To find the second derivative, we will differentiate f'(x) using the chain rule.
Let's denote f'(x) as g(x) for simplicity.
g(x) = (2 - e^x) * e^2x
g'(x) = d/dx((2 - e^x) * e^2x)
= (2 - e^x) * d/dx(e^2x) + e^2x * d/dx(2 - e^x)
= (2 - e^x) * 2e^2x + e^2x * (-e^x)
= 4e^2x - 2e^3x - e^3x
= 4e^2x - 3e^3x
The second derivative of the function f(x) = ln(e^x - 1 / e^x) is f''(x) = 4e^2x - 3e^3x.
Therefore, the derivative of the given function is f'(x) = (2 - e^x) * e^2x, and the second derivative is f''(x) = 4e^2x - 3e^3x.