Calculus derivative...easy one i think i just cant figure it out lol?

y=log(4x) suppose u=4x

Then,

y=log(u)----------(1) and u=4x-------(2)

Differentiating equation 1 with respect to u and equation 2 with respect to x we get,

dy/du=1/u -------(3) and du/dx=4------(4)

Multiplying 3and 4 we get,

dy/dx=4/u

Replacing u by 4x we get,

dy/dx=1/x (Ans)

First, use the laws of logarithms to write this as log 4 + log x. Now, log 4 is a constant, and depending on whether your text uses log to mean natural logarithm (some do) or common logarithm (i.e. log base 10), you should now be able to use a formula from your text (which you should memorize for your exam).

In the worst case, log means common logarithm and you were only give the derivative formula for natural logarithms [ln x]' = 1/x . (Using the other notation for natural log to be clear). In that case, you'd need to use the change of base formula for logarithms before differentiating.

Incidentally, you could also do this by using the derivative formula for whichever kind of logarithm your text means and the chain rule (since you plugged 4x into the logarithm function).