Question

$Differentiate \sf {x^{sin\ x}}xsin x , x > 0 w.r.t.x$
please help me fast....
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Answer 1

Answer:-

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

➽ Taking log Both Side:-

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

Distance is the total movement of an object without any regard to direction. We can define distance as to how much ground an object has covered despite its starting or ending point.