Solve this difficult derivative...?

Question

y = (lnx)/(ln3). Find y'.

The answer is 1/(xln3). Can someone show the work and rules used?

Thanks!

Ding ding ding!

Anan T = winner!

just what i was looking for!!... it was a constant... thanks again!

Answer 1

y = (1/ ln 3) (ln x) = A (ln x)
dy/dx = A (1 / x)
dy/dx = (1 / ln 3) (1 / x)
dy/dx = 1 / [ (ln 3)(x) ]

Answer 2

Well I'm pretty sure you've learned the quotient rule for taking derivatives. SO if you have a function y = f(x)/g(x), its the bottom times the derivative of the top (g(x) X f'(x)) minus the top times derivative of the bottom (f(x) X g'(x)) and put all of that over the bottom squared ((g(x))^2). So remember in your problem the derivative of (lnx) is 1/x and the derivative of (ln3) is zero since its a constant. So bottom times derivative of the top is (ln3 X 1/x) then the top times derivative of the bottom is zero. The bottom squared is (ln3)^2. Therefore you have (ln3 X 1/x)/(ln3)^2. An (ln3) cancels on the top and bottom and u get 1/(xln3)

Answer 3

This is acually not a hard one at all. a
ln(x)/ln(3) =
1/ln(3) * ln(x) ...1/ln(x) is just a constant so you can ignore it when taking the derivatives. now all u need is just to take the derivative of ln(3). which is 1/x
so the answer will be 1/ln(3) *1/x
which equals 1/(x*ln3). hope this helps.

Anan Tak

Answer 4

y=(lnx) / (ln3)
you will use the quotient rule
dy/dx= V'U - UV' / V²
V=ln3 and U = lnx
dy/dx = (ln3)(1/x) / (ln3)²
dy/dx = ln3/x / (ln3)²
dy/dx = 1/x(ln3)