find the max value of f''(x) f(x)=1/(x^2+1) interval [0,3]
do i have to find the derivative first?
i guess you have to find the derivative twice
yes, the '' part, correct?
but it is not hard if you write it as \[f(x)=(x^2+1)^{-1}\] and use the chain rule
a tutor mentioned when there is two '' signs in the f(x) function, i have to find the derivative 3 times, what do you think?
yes
first derivative is \[-2x(x^2+1)^{-2}\]
no i do not think so
No that is the second derivative. You only need to take the derivative twice
first derivative if \(f'\) second derivative is \(f''\)
yes, i got that for my first derivative. for my second derivative i got, -4x(x^2+1)^-3
this is because u have to check where the third derivative of function goes zeri so that such point where it goes zero will denote point of extremum of second derivative
ooooh i see hold the phone to answer the question above, find the max of \(f''\) then yes, you need the third derivative to find the max of the second derivative
what @thinker said
yay :D
so! onto the second derivative!
we use the product rule to find the second derivative, right?
i have my work written all over the place so i have to redo my problem from yesterday
but first you need the second derivative, which i think is \[\frac{8x}{(x^2+1)^3}\]
i would write the first derivative as \[\frac{-2x}{(x^2+1)^2}\] and then take the derivative of that
you are going to have to add them up anyway, so rather than use the product rule it is going to be best to use the quotient rule, then you don't have to add the expressions, you already have it as a quotient your final job is going to be to find the zeros of the third derivative, so you need it in fraction form
ok, so i'll use the quotient rule, brb while i do this
ok do some algebra to simplify also, and you can check it with the answer i wrote above
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