If y= e^x - e^-x / e^x + e^-x , then y'=
Please help
I took the derivative and got
Y'= (e^x+ e^-x) ( e^x + e^-x) - (e^x + e^-x) (e^x - e^-x)/ (e^x + e^-x)^2
You will need brackets
y= (e^x - e^-x) / (e^x + e^-x)
wonder what would happen if we multiply top and bottom by e^x
= y= (e^x - e^-x) / (e^x + e^-x) * (e^x/e^x)
= (e^2x - 1)/(e^2x + 1)
now dy/dx
= ( (e^2x + 1)(2e^2x) - (e^2x - 1)(2e^2x) )/(e^2x + 1)^2
= (2e^4x + 2e^2x - 2e^2x + 2e^4x)/(2e^2x + 1)^2
= 4e^4x/(2e^2x + 1)^2
check my work, I should have written it out on paper first
I know the correct answer is 4 / (e^x + e^-x)^2
You should also know that
y = tanh(x)
y' = sech^2(x) = [2/(e^x+e^-x)]^2
To find the derivative of y = (e^x - e^-x) / (e^x + e^-x), we can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x) / h(x), where both g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's use this rule to find the derivative of y.
First, let's find the derivatives of the numerator and denominator separately:
- The derivative of e^x is e^x, since the derivative of e^u is e^u * u' (where u is a function of x and u' is the derivative of u with respect to x).
- The derivative of e^-x is -e^-x, using the same logic.
Next, let's find the derivative of the numerator:
g'(x) = (e^x - (-e^-x)) = e^x + e^-x
Now, let's find the derivative of the denominator:
h'(x) = (e^x + e^-x)
Finally, let's substitute our values into the quotient rule formula:
y' = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= ((e^x + e^-x) * (e^x + e^-x) - (e^x - e^-x) * (e^x + e^-x)) / ((e^x + e^-x))^2
= (e^(2x) + 2 + e^(-2x) - (e^(2x) - 2 + e^(-2x))) / ((e^x + e^-x))^2
= (4) / ((e^x + e^-x))^2
Therefore, the derivative y' of y = (e^x - e^-x) / (e^x + e^-x) is y' = 4 / ((e^x + e^-x))^2.