Question

Differentiate $\sf {x^{sin\ x}}$ , x > 0 w.r.t.x

Answer 1

Answer:

$\sf {\underline{\bigstar\ Let\ y\ =\ x^{sin\ x}}}$

➽ Taking log both sides :-

$\sf \mapsto {log\ y\ =\ log\ x^{sin\ x}}$

$\sf \mapsto {log\ y\ =\ sin\ x\ .\ log\ x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf {[log\ a^b\ =\ b\ log\ a]}$

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

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

$\: \: \: \: \: \: \: \: \: \begin{gathered} \small\boxed { \begin{array}{cc} \sf {By\ product\ rule} \\ \\ \sf {(uv)'\ =\ u'v\ +\ v'u} \\ \\ \sf {Where\ u\ =\ sin\ x\ and\ v\ =\ log\ x} \end{array} } \end{gathered}$

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

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

$\sf : \implies {\dfrac{dy}{dx}\ \dfrac{1}{y}\ =\ cos\ x\ log\ x\ +\ sin\ x\ \dfrac{1}{x}}$

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

$\sf {\underline{\bigstar\ Putting\ back\ y\ =\ x^{sin\ x}}}$

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

$\sf : \implies {x^{sin\ x}\ cos\ log\ x\ +\ x^{sin\ x}\ \dfrac{1}{x}sin\ x}$

$\sf : \implies {x^{sin\ x}\ cos\ log\ x\ +\ x^{sin\ x}\ x^{-1}\ sin\ x}$

${\therefore{\underline{\boxed{\sf {x^{sin\ x}\ cos\ log\ x\ +\ x^{sin\ x\ -\ 1}\ sin\ x}}}}}$