If f(x) = lnx / (x+2) what is f'(1) ?
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calendarEducator since 2010
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f(x) = ln x / (x + 2)
f'(x) = [(ln x)' *(x + 2) - ln x*(x + 2)']/(x + 2)^2
=> f'(x) = [1/x*(x + 2) - ln x] / (x + 2)^2
For x = 1
f'(1) = (3 - ln 1) / 3^2
=> 1/3
The value of f'(1) = 1/3
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calendarEducator since 2008
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f(x) = lnx/(x+2)
We need to find the values of f'(1)
First we will need to find the first derivative f'(x).
Let f(x) = u/v such that:
u= ln x ==> u' = 1/x
v= x+2 ==> v' = 1
==> f'(x) = u'v-uv' / v^2
==> f'(x) = [ (1/x)*(x+2) - lnx] / (x+2)^2
==> f'(x) = (x+2)/x - ln x] / (x+2)^2
==> Now we will substitute with x = 1
==> f'(1) = (1+2)/1 - ln 1] / (1+2)^2
= (3 - 0) / 9 = 3/9 = 1/3
==> f'(1)= 1/3