f'(x).... y=(e^x-e^-x)/(e^x+e^-x). I need the calculation work.
[ans: 4/(e^x+e^-x)^2]
I'm not exactly sure what your asking for or what the question is? I assume your trying to solve for the derivative? Did you try plugging it into wolfram alpha an clicking on show steps? That always works. Well not always, but it defiantly well if interperting (e^x-e^-x)/(e^x+e^-x) this correctly to be the original function
f(x) = (e^x - 1/e^x) / (e^x + 1/e^x))
how about multiplying each term by e^x to get
f(x) = (e^2x - 1)/(e^2x + 1)
f'(x) = [(e^2x + 1)(2e^2x) - (e^2x - 1)(2e^x)]/(e^2x + 1)^2
= 4e^(2x) / (e^2x + 1)^2
To find the derivative of y with respect to x, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then the derivative of f(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply this rule to the given function y = (e^x - e^(-x)) / (e^x + e^(-x)):
Step 1: Find the derivative of the numerator g(x) = e^x - e^(-x)
- The derivative of e^x is e^x.
- The derivative of -e^(-x) is e^(-x).
Therefore, g'(x) = e^x - e^(-x)
Step 2: Find the derivative of the denominator h(x) = e^x + e^(-x)
- The derivative of e^x is e^x.
- The derivative of e^(-x) is -e^(-x).
Therefore, h'(x) = e^x - e^(-x)
Step 3: Substitute the values into the quotient rule formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Plugging in the values we found:
f'(x) = ((e^x - e^(-x)) * (e^x + e^(-x)) - (e^x - e^(-x)) * (e^x - e^(-x))) / ((e^x + e^(-x))^2)
Simplifying the expression further:
f'(x) = (e^2x + e^(-2x) - (e^2x - e^(-2x))) / ((e^x + e^(-x))^2)
f'(x) = (2e^(-2x)) / ((e^x + e^(-x))^2)
Therefore, the derivative of y with respect to x, f'(x), is:
f'(x) = 2e^(-2x) / (e^x + e^(-x))^2
Finally, we can simplify the expression further using properties of exponents:
f'(x) = 2 / (e^x + e^(-x))^2
So, the calculation work shows that the derivative of y = (e^x - e^(-x)) / (e^x + e^(-x)) with respect to x is equal to 2 / (e^x + e^(-x))^2, which matches the provided answer.