Dy/dx of y=ln2(log base2 x)

Why is the answer 1/x ? Intuitively I would think that i would have to use the product rule:

[(ln2)((dy/dx)(log base2 x))] [(log base2 x) ((dy/dx)(ln2))]

[ln2/(xln2)][(log base2 x)/2]
...
The answer key leaves the ln2 and only takes the derivative of the log base2 x. Why? Thanks

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y = ln2*(log₂x)         ← Note: The formula for the derivative of a logarithm is written
                                           specifically for only natural logarithms ... So, we must
                                           convert the base 2 logarithm to natural logarithms

y = ln2*(lnx / ln2)       ← Now, the ln2's cancel
y = lnx                       ← Now, take the derivative

dy/dx = 1/x                       ← ANSWER


Have a good one!
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In this case Ln 2 is a constant. Also log base 2 of x = ln x / ln 2. Then:

y ' = ( ln 2) ( 1/ x) ( 1/ ln 2)................... Derivative of ln x/ ln 2 = ( 1/ x. ln 2 )

After canceling the ln 2 we have : y ' = 1/ x........ THE ANSWER.

Note : If F(x) = k. f(x) ... then ..... F ' (x) = k f ' (x). In this case k = ln 2

ln(2) is a constant, so

dy/dx = ln(2) * d/dx(log(base 2)(x)) = ln(2) *(log(base 2)(e) / x) = 1/x, because

log(base 2)(e) = 1/ln(2), (change of base rule).

Note that for a > 0, d/dx(log(base a)(x)) = log(base a)(e) / x.