# Show how prove f' increase if f=x-ln(e to x +1)?f function

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You need to find the second derivative of the function. Since the second derivative is the derivative of the first derivative of the function, hence you need to find the first derivative of the function.

`f'(x) = (x-ln(e^x +1))' =gt f'(x) = 1 - (e^x +1)'/(e^x +1)`

`f'(x) = 1 - e^x/(e^x +1) =gtf'(x) =(e^x - e^x +1)/(e^x +1)`

`f'(x) = 1/(e^x +1)`

You need to differentiate the first derivative with respect to x such that:

`f"(x) = (1'*(e^x + 1) - 1*(e^x +1)')/(e^x + 1)^2`

`f"(x) = (0*(e^x + 1) - 1*(e^x))/(e^x + 1)^2`

`f"(x) = -e^x/(e^x + 1)^2`

Since the denominator is positive all the time, the negative numerator makes the fraction negative => f"(x)<0.

**Hence, the first derivative decreases because f"(x)<0.**