amonoconnor
  • amonoconnor
Would someone be willing to check my work for this one? "Find [d^2f/dx^2](x) if f(x) = log(base2)x." Any and all help is greatly appreciated!
Calculus1
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SOLVED
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
i'd switch it out of base 2 before even starting...
amonoconnor
  • amonoconnor
Here's what I did: \[f(x) = \log_{2}x \] \[f'(x) = \frac{1}{x* \ln 2}\] \[f''(x) = \frac{(0)*(xln2) - 1*[(1)*\ln2 + x(1/2)]}{(xln2)^2}\]
amonoconnor
  • amonoconnor
I used the formula for the derivative of a log with an arbitrary base for f'(x), and Quotient Rule for f''(x).

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amonoconnor
  • amonoconnor
Is this the correct final answer? \[f''(x) = \frac{-\frac{1}{2}x - \ln2}{(xln2)^2}\]
IrishBoy123
  • IrishBoy123
no, it's much more straightforward than that
amonoconnor
  • amonoconnor
:/ Rgg... Would you walk me through it?
IrishBoy123
  • IrishBoy123
\(f'(x) = \dfrac{1}{x* \ln 2} = \dfrac{1}{\ln 2} . \dfrac{1}{x}\) what you want for the second deriv is \((\dfrac{1}{x})'\)
IrishBoy123
  • IrishBoy123
\[\dfrac{1}{x} = x^{-1}\] you can use that rule \((x^n)' = n x^{n-1}\)
amonoconnor
  • amonoconnor
Then what happens to the 1/ln2?
IrishBoy123
  • IrishBoy123
treat \(\dfrac{1}{\ln 2} \) as a constant because it is a constant...
amonoconnor
  • amonoconnor
So it's like using the Product Rule, and 1/ln2 just stays there?
IrishBoy123
  • IrishBoy123
if you want to see it that way but i think that is still overcomplicating it... think: \( (Ax^{n})' = n A x^{n-1} \)
IrishBoy123
  • IrishBoy123
just as \((A \sin x)' = A \cos x\) \( (A e^{x})' = A e^x \) \((A \ln x)' = \frac{A}{x}\) constant just hangs there....but yes you could use a product rule if you wanted too but it would be unnecessary....
amonoconnor
  • amonoconnor
Is this the final answer then? \[f''(x) = \frac{1}{\ln 2}*(-1x^{-2})\] \[= \frac{1}{\ln2}*\frac{-1}{x^{-2}}\] \[= \frac{-1}{(x^{-2})\ln2}\] ?
amonoconnor
  • amonoconnor
pellet, I see a mistake. Take the negative from the x^-2 after I put it in the denominator!! My bad!
amonoconnor
  • amonoconnor
Thank you:)
IrishBoy123
  • IrishBoy123
mp constants flow through like this... \(\large f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\) \(\large A f'(a)=\lim_{h\to 0}\frac{Af(a+h)-Af(a)}{h}\)
amonoconnor
  • amonoconnor
I know how constants work, I just didn't know that ln(a) would be a constant, and not have to have its derivative taken again. :)
IrishBoy123
  • IrishBoy123
great!

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