How do you use logarithmic differentiation to find the derivative of the function?

1 Answer
Mar 28, 2015

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