# Logarithmic Differentiation

If $y = f\left( x \right)$ is a complicated function, i.e. it involves several products of functions, quotients or radical signs, then we take logarithms on both sides which will make differentiation much easier. Such differentiation is called logarithmic differentiation.

Thus, taking logarithms on both sides of the given equation, we have
$\ln y = \ln f\left( x \right)$

Differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{d}{{dx}}\ln y = \frac{d}{{dx}}\ln f\left( x \right) \\ \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{f\left( x \right)}}f’\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{y}{{f\left( x \right)}}f’\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{f\left( x \right)}}{{f\left( x \right)}}f’\left( x \right),\,\,\,\,y = f\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = f’\left( x \right) \\ \end{gathered}$

This shows that the result is the same as what we would have without taking $\ln$.

Example: Differentiate $\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{2/3}}}}{{{{\left( {2 – 3x} \right)}^{3/4}}{{\left( {3 – 4x} \right)}^{4/3}}}}$ with respect to $x$.

Consider the function $y = \frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}$, then taking $\ln$ both sides, we get
$\begin{gathered} \ln y = \ln \left[ {\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}} \right] \\ \Rightarrow \ln y = \ln \sqrt x + \ln {\left( {1 – 2x} \right)^{\frac{2}{3}}} – \ln {\left( {2 – 3x} \right)^{\frac{3}{4}}} – \ln {\left( {3 – 4x} \right)^{\frac{4}{3}}} \\ \Rightarrow \ln y = \frac{1}{2}\ln x + \frac{2}{3}\ln \left( {1 – 2x} \right) – \frac{3}{4}\ln \left( {2 – 3x} \right) – \frac{4}{3}\ln \left( {3 – 4x} \right) \\ \end{gathered}$

Differenting both sides with respect to $x$, we have
$\begin{gathered} \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = y\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}}} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = \left[ {\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}} \right]\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}}} \right] \\ \end{gathered}$