The derivative of natural log?

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

2 Answers
Jun 18, 2018

#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)#

Jun 18, 2018

#-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!