Find the derivative of the function y = lnx/1+lnx?

Wednesday, March 5, 2014

Find the derivative of the function y = lnx/1+lnx?

Since the function is a quotient, we'll use the quotient
rule to differentiate the function with respect to
x:

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{\left({\ln{{x}}}\right)}'\cdot{\left({1}+{\ln{{x}}}\right)}-{\left({\ln{{x}}}\right)}{\left({1}+{\ln{{x}}}\right)}'}}{{{1}+{\ln{}}}}$
x)^2

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\frac{{1}}{{x}}+\frac{{{\ln{{x}}}}}{{x}}-\frac{{{\ln{{x}}}}}{{x}}}}{{{1}+{\ln{}}}}$
x)^2

We'll reduce like terms from
numerator:

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\frac{{1}}{{x}}}}{{{1}+{\ln{}}}}$
x)^2

Therefore, the requested derivative of
the function is

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{{x}\cdot{\left({1}+{\ln{}}\right.}}}$
x)^2)

Help Notes

Wednesday, March 5, 2014