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Derivative, where to go from here!?
Have to find first and second derivative:
Started with:
f(x) = (ln(x)) / (1+e^x)
Used quotient rule and got:
[(1+e^x)1/x - ln(x)(e^x)] / (1+e^x)^2
Just need the next step, not the whole problem. Just not sure where to go from here. Any help would be greatly appreciated!
2 Answers
- 7 years ago
Do the quotient rule again. I suggest doinf the derivatives seperately and then plugging in. Do the entire numerator first. The 1 in the numerator of the first term is not needed.
Derivative is (xe^x-e^x-1))/x^2.
For the second term, use product rule.
It is lnxe^x+e^x/x= e^x(lnx+1/x)
So the derivative of the entire numerator is
((xe^x-e^x-1))/x^2)-(e^x(lnx+1/x)).
The derivative of the denominator is:
2(1+e^x)*e^x=2e^x(1+e^x)=2e^x+2e^2x
Now plug in and simplify, careful with mistakes as there is a lot of algebra.
- ?Lv 77 years ago
y ' = [(1+e^x)1/x - ln(x)(e^x)] / (1+e^x)^2
u = [(1+e^x)1/x - ln(x)(e^x)] --->u' = e^x /x - (1/x^2)(1+e^x) - e^x / x - ln(x)e^(x) = - (1/x^2)(1+e^x) - ln(x)e^(x)
v=(1+e^x)^2 --->v ' =2e^x(1+e^x)
and use the quotient rule to find y "