9.0 Logarathimic Differentiation

  Differentiation

$$y = {\left( {f\left( x \right)} \right)^{g\left( x \right)}}$$
To avoid product and quotient rule, we can use this method

Step 1: Taking $\log$ on both side we get,
$$\log y = \log \left( {{{\left( {f\left( x \right)} \right)}^{g\left( x \right)}}} \right)$$Step 2: Using property of log, we get, $$\log y = g\left( x \right)\log f\left( x \right)$$Step 3: We will differentiate the above function, then will substitute the value of $y$ in the final answer.

Question 1: Find derivative of $y = {\left( {\sin x} \right)^{{x^3}}}$

Solution: $$ y = {\left( {\sin x} \right)^{{x^3}}} $$ Taking $\log$ on both side, $$ \log y = \log \left( {{{\left( {\sin x} \right)}^{{x^3}}}} \right) $$$$ \log y = {x^3}\log \left( {\sin x} \right) $$ On differentiating we know that, $$ \frac{d}{{dx}}\log x = \frac{1}{x} $$We get,$$ \frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{x^3}\log \left( {\sin x} \right)} \right) $$Using product rule, $$ \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {\sin x} \right)\frac{d}{{dx}}{x^3} + {x^3}\frac{d}{{dx}}\left( {\log \left( {\sin x} \right)} \right) $$We know that, $$ \frac{d}{{dx}}{x^n} = n{x^{n - 1}} $$$$ \frac{d}{{dx}}\sin x = \cos x $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {\sin x} \right)3{x^{3 - 1}} + {x^3}\frac{1}{{\sin x}}\frac{d}{{dx}}\left( {\sin x} \right) $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = 3{x^2}\log \left( {\sin x} \right) + {x^3}\frac{{\cos x}}{{\sin x}} $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = {x^2}\left( {3\log \left( {\sin x} \right) + x\cot x} \right) $$$$ \frac{{dy}}{{dx}} = y{x^2}\left( {3\log \left( {\sin x} \right) + x\cot x} \right) $$Substitute value of $y$, $$ \frac{{dy}}{{dx}} = {\left( {\sin x} \right)^{{x^3}}}{x^2}\left( {3\log \left( {\sin x} \right) + x\cot x} \right) $$

Question 2: Find derivative of $y = {x^x}$

Solution: Taking $\log$ on both side, $$ \log y = \log \left( {{x^x}} \right) $$$$ \log y = x\log \left( x \right) $$On differentiating we know that, $$ \frac{d}{{dx}}\log x = \frac{1}{x} $$
We get,
$$ \frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {x\log \left( x \right)} \right) $$
Using product rule,
$$ \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( x \right)\frac{d}{{dx}}x + x\frac{d}{{dx}}\left( {\log \left( x \right)} \right) $$
We know that,
$$ \frac{d}{{dx}}{x^n} = n{x^{n - 1}} $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( x \right){x^{1 - 1}} + x\frac{1}{x}\frac{d}{{dx}}\left( x \right) $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = \log x + {x^{1 - 1}} $$$$ \frac{1}{y}\frac{{dy}}{{dx}} = \left( {\log x + 1} \right) $$$$ \frac{{dy}}{{dx}} = y\left( {\log x + 1} \right) $$

Substitute value of $y$,
$$ \frac{{dy}}{{dx}} = {x^x}\left( {\log x + 1} \right) $$