Question

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!

Answer 1

i'd switch it out of base 2 before even starting...

Answer 2

Here's what I did:

\[f(x) = \log_{2}x \]
\[f'(x) = \frac{1}{x* \ln 2}\]
\[f''(x) = \frac{(0)*(xln2) - 1*[(1)*\ln2 + x(1/2)]}{(xln2)^2}\]

Answer 3

I used the formula for the derivative of a log with an arbitrary base for f'(x), and Quotient Rule for f''(x).

Answer 4

Is this the correct final answer?
\[f''(x) = \frac{-\frac{1}{2}x - \ln2}{(xln2)^2}\]

Answer 5

no, it's much more straightforward than that

Answer 6

:/ Rgg...
Would you walk me through it?

Answer 7

\(f'(x) = \dfrac{1}{x* \ln 2} = \dfrac{1}{\ln 2} . \dfrac{1}{x}\)

what you want for the second deriv is \((\dfrac{1}{x})'\)

Answer 8

\[\dfrac{1}{x} = x^{-1}\]

you can use that rule \((x^n)' = n x^{n-1}\)

Answer 9

Then what happens to the 1/ln2?

Answer 10

treat \(\dfrac{1}{\ln 2} \) as a constant because it is a constant...

Answer 11

So it's like using the Product Rule, and 1/ln2 just stays there?

Answer 12

if you want to see it that way but i think that is still overcomplicating it...
think:
\( (Ax^{n})' = n A x^{n-1} \)

Answer 13

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....

Answer 14

Is this the final answer then?
\[f''(x) = \frac{1}{\ln 2}*(-1x^{-2})\]
\[= \frac{1}{\ln2}*\frac{-1}{x^{-2}}\]
\[= \frac{-1}{(x^{-2})\ln2}\]
?

Answer 15

pellet, I see a mistake. Take the negative from the x^-2 after I put it in the denominator!! My bad!

Answer 16

Thank you:)

Answer 17

mp

constants flow through like this...

\(\large f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\)

\(\large A f'(a)=\lim_{h\to 0}\frac{Af(a+h)-Af(a)}{h}\)

Answer 18

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. :)

Answer 19

great!