Find the derivative of
y=ln((1+e^x)/(1-e^x))
I know that the derivative of y=ln(1+e^x) would be (e^x)/(1+e^x), but I'm not sure what to do with the 1-e^x on the bottom here. Can someone help please? Thanks!
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wolfram alpha
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Wolfram Alpha:Computational Knowledge Engine
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derivartive ln((1+e^x)/(1-e^x))
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Then clic option Show steps
first of all recall that
ln (A/B) = ln A - ln B
so
y = ln((1+e^x)/(1-e^x) )
= ln (1+e^x) - ln (1-e^x)
dy/dx = e^x/(1+e^x) - (-e^x)/(1 - e^x)
simplify as needed
Okay, thanks guys!
To find the derivative of the given function, we can use the quotient rule. The quotient rule states that the derivative of the quotient of two functions, u(x)/v(x), is given by:
[y'(x) = (u'(x)v(x) - u(x)v'(x))/[v(x)]^2]
In this case, u(x) = ln((1+e^x), and v(x) = (1-e^x). Let's find the derivatives of u(x) and v(x) first.
1. Derivative of u(x) = ln(1+e^x):
To find the derivative of ln(1+e^x), we will apply the chain rule. Let's define g(x) = 1+e^x, then we have u(x) = ln(g(x)), where g(x) is inside the natural logarithm function.
(u'(x)) = (1/g(x)) * g'(x)
Apply the chain rule:
g'(x) = (e^x) * 1 = e^x
Substitute the values back into (u'(x)) = (1/g(x)) * g'(x):
(u'(x)) = (1/(1+e^x)) * (e^x)
Simplifying, we have:
(u'(x)) = (e^x)/(1+e^x)
2. Derivative of v(x) = (1 - e^x):
The derivative of v(x) = (1 - e^x) with respect to x is straightforward:
(v'(x)) = -e^x
Now that we have the derivatives of u(x) and v(x), we can use the quotient rule formula to solve for the derivative of y(x).
[y'(x) = (u'(x)v(x) - u(x)v'(x))/[v(x)]^2]
= ([(e^x)/(1+e^x)] * (1 - e^x) - ln(1+e^x) * (-e^x))/[(1 - e^x)]^2
Simplifying the expression further will give us the final derivative of y(x).