Question

The derivative of natural log?

`y=-ln (3/2x)`

Answer 1

`y' = - frac(1)(x)`

Explanation:

We have: `y = - ln(frac(3)(2) x)`

We must use the chain rule to differentiate `y`.

Let `u = frac(3)(2) x Rightarrow u' = frac(3)(2)` and `v = - ln(u) Rightarrow v' = - frac(1)(u)`:

`Rightarrow y' = u' cdot v'`

`Rightarrow y' = frac(3)(2) cdot - frac(1)(frac(3)(2) x)`

`Rightarrow y' = frac(3)(2) cdot - frac(2)(3 x)`

`therefore y' = - frac(1)(x)`

Answer 2

`-1/x`

Explanation:

Remember the form:

`(d)/(dx)lnf(x) = f^'(x)/f(x)`

`y = -ln((3x)/2)`

`y = ln(2/(3x))`

`\therefore(dy)/(dx) = -2/(3x^2) \div 2/(3x)`

`(dy)/(dx) = -1/x`

Hope that makes sense!