Find the second derivative and it's critical points of f(x)= x/x^2+2
The critical point is where the derivative of f(x)=0 so u need to differentiate once to solve the second part and twice for the first part of ur question. Are u familiar with differenitating a polynominal?
Yeah I did the first derivative but then when I did the 2nd it got alittle messy
Is f(x)=x/(x^2+2) or =x/x^2 +2? It will get messy if it is the later.
It's x/x^2 +2
Then u can simplify 1st be4 differenitate - what would be the simplified form?
1/x+2?
That is correct - so now it should be easier to differeniate ;)
Hmm okay let me see what I get
Post back if u run into any trouble or need more help :)
Okay so I took the derivative and got -x+2/(x+2)^2 idk if that's right
first derivative should be much simplier than that - plz try again or refer to ur book?
f(x)=1/x+2....so what is d/dx(1/x) + d/dx(2)?
Hmm wouldn't i do a quotient rule?
Qutient rule to perform d/dx(1/x) and addition rule for d/dx(A+B)= dA/dx + dB/dx
Wait , omg I just realized I wrote down the wrong problem it's x/(x^2+2) >.< sorry
:( yup then ur life just got a bit more messy now >+<
Ah man I know :/ no wonder I was so confused lol
OK so quotient rule indeed: d/dx(u/v) = (vdu/dx-udv/dx)/dx... It will be quite long....oh let me see if u got it already....hold on....
f'(x) should be equal to (-x^2+2)/(x^2+2)^2 plz see if u get the same thing :)
Okay I get (-x^2 +2/)(x^2+2)^2^-^
So the critical points for the numerator will be positive and negative square root of 2 ?
yup yup n thats answer to the 2nd part of ur Q...first part: more differentiation :(
Okay so for the second part of when I find f ''(x) can I make it (x^2+2) ( x^2 -2) ^ -2 and do the product rule. ?
Nope quotient rule again because u are missing a "/" there...its a pain but let me know if u wanna try it ;)
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