derivative of f(y)=yln(1+e^y)
To find the derivative of the function f(y) = yln(1+e^y), we can use the product rule and chain rule.
Let's start by applying the product rule to differentiate the function:
f'(y) = (y)'ln(1+e^y) + y(ln(1+e^y))'
Now, let's find the derivative of each term.
The derivative of y with respect to y is simply 1:
(y)' = 1
Next, let's find the derivative of ln(1+e^y) using the chain rule. The chain rule states that if we have a function of the form f(g(y)), then the derivative of f(g(y)) with respect to y is f'(g(y)) * g'(y).
In this case, f(u) = ln(u) and g(y) = 1 + e^y. So, we can write:
(ln(1+e^y))' = f'(g(y)) * g'(y)
To find f'(u), we differentiate ln(u) with respect to u, which gives us 1/u:
f'(u) = 1/u
Now, let's find g'(y) by differentiating 1+e^y with respect to y:
g'(y) = (1)' + (e^y)'
The derivative of 1 with respect to y is 0, and the derivative of e^y with respect to y is simply e^y:
g'(y) = 0 + e^y = e^y
Now that we have f'(u) and g'(y), let's substitute them back into the chain rule:
(ln(1+e^y))' = f'(g(y)) * g'(y) = (1 / (1+e^y)) * e^y
Finally, substituting the derivatives we found earlier back into the original product rule equation:
f'(y) = (y)'ln(1+e^y) + y(ln(1+e^y))'
= 1 * ln(1+e^y) + y(1 / (1+e^y)) * e^y
Simplifying further, we get:
f'(y) = ln(1+e^y) + ye^y / (1+e^y)
Therefore, the derivative of f(y) = yln(1+e^y) is ln(1+e^y) + ye^y / (1+e^y).