Anyone can help me please with this: dy/dx where y=x ln 2x.

Question

Please can someone help me with this dy by dx question, for calculation

Answer 1

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) = 1

 

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