Logarithmic differentiation when the logarithm of an expression can be written in a form for which we can find the derivative more simply. Or, for which we can find the derivative at all.

Here's one of each:

**Example 1**

For #y=(3x+5)^5/(4x-9)^7#, we could find #(dy)/(dx)# by using the quotient rule, the power rule and the chain rule and then simplify algebraically.

OR

#ln y=ln((3x+5)^5/(4x-9)^7)=5ln(3x+5)-7ln(4x-9)#

Differentiating implicitly gives us:

#1/y (dy)/(dx) = 5(1/(3x+5)*3)-7(1/(4x-9)*4)#.

#1/y (dy)/(dx) = 15/(3x+5)- 28/(4x-9)#

Remembering that #y=(3x+5)^5/(4x-9)^7#, we can solve for #(dy)/(dx)#

#(dy)/(dx) = y(15/(3x+5)- 28/(4x-9))#

#(dy)/(dx) = (3x+5)^5/(4x-9)^7(15/(3x+5)- 28/(4x-9))#

**Example 2**

#y= root(x)x=x^(1/x)#.

This is neither an exponential, nor a power function. we need some other way to find #(dy)/(dx)#. logarithmic differentiation will work.

#ln y = 1/x lnx#

#1/y (dy)/(dx) = -1/x^2 lnx + 1/x * 1/x=(1-lnx)/x^2#

Solving gives:

#(dy)/(dx) =( root(x)x(1-lnx))/x^2#