Question

Differentiate w.r.t.x the function,

(log x)^log x , x > 1

Answer 1

Step-by-step explanation:

$\sf {\underline{\bigstar Let\ y\ =\ (log\ x)^{log\ x}}}$

➽ Taking log both sides :-

$\sf \mapsto {log\ y\ =\ log\ \bigg((log\ x)^{log\ x} \bigg)}$

$\sf \mapsto {log\ y\ =\ log\ x.\ log\ (log\ x)}$

$\sf {\underline{\bigstar Differentiating\ both\ sides\ w.r.t.x}}$

$\sf : \implies {\dfrac{d(log\ y)}{dx}\ =\ \dfrac{d \big(log\ x.\ log\ (log\ x) \big)}{dx}}$

$\sf : \implies {\dfrac{d(log\ y)}{dx} \bigg(\dfrac{dy}{dy} \bigg)\ =\ \dfrac{d \big(log\ x.\ log\ (log\ x) \big)}{dx}}$

$\sf : \implies {\dfrac{d(log\ y)}{dy} \bigg(\dfrac{dy}{dx} \bigg)\ =\ \dfrac{d \big(log\ x.\ log\ (log\ x) \big)}{dx}}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{d \big(log\ x.\ log\ (log\ x) \big)}{dx}}$

$\begin{gathered} \small\boxed{ \begin{array}{cc} \sf{Using\ product\ rule\ in\ log\ x\ .log\ (log\ x)} \\ \\ \sf {As\ (uv)'\ =\ u'v\ +\ v'u} \\ \\ \sf {Where,\ u\ =\ log\ x\ and\ v\ =\ log\ (log\ x)} \end{array} } \end{gathered}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{d(log\ x)}{dx}\ .\ log\ (log\ x)\ +\ \dfrac{d \big(log\ (log\ x) \big)}{dx}\ .\ log\ x}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{1}{x}\ log\ (log\ x)\ +\ \dfrac{1}{log\ x}\ .\ \dfrac{d(log\ x)}{dx}\ .\ log\ x}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{1}{x}\ log\ (log\ x)\ +\ \dfrac{d(log\ x)}{dx}\ .\ \dfrac{log\ x}{log\ x}}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{1}{x}\ log\ (log\ x)\ +\ \dfrac{1}{x}}$

$\sf : \implies {\dfrac{1}{y}\ .\ \dfrac{dy}{dx}\ =\ \dfrac{1}{x}\ log\ (log\ x)\ +\ \dfrac{1}{x}}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ y \bigg(\dfrac{1}{x}\ +\ \dfrac{log\ (log\ x)}{x} \bigg)}$

$\sf \therefore {\underline{\boxed{\sf \dfrac{dy}{dx}\ =\ (log\ x)^{log\ x}\ \bigg(\dfrac{1}{x}\ +\ \dfrac{log\ (log\ x)}{x} \bigg)}}}$