Jon P. answered • 09/21/15

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You'll need to use two rules: derivative of a product, and chain rule.

Let's let f(x) = x, and g(x) = ln 2x

D

_{x}f(x) g(x) = f(x) D_{x}g(x) + g(x) D_{x}f(x)D

_{x}g(x) = 1/(2x) D_{x}2x = 1/(2x) * 2 = 1/x(As an aside, notice that this is the same as the derivative of ln x. That may seem odd, but remember, ln 2x = ln 2 + ln x. Since ln 2 is a constant, the derivative of ln 2 + ln x is 0 + 1/x = 1/x. So the derivative of the ln of ANY constant multiple of x is just 1/x.)

D

_{x}f(x) = 1So the derivative of the original expression is: x (1/x) + ln 2x * 1 = 1 + ln 2x