Question

Find f'(x) of f(x)= sin{x/(x - sin[x/(x-sinx)])}

Answer 1

Solution:

$f(x)=\sin\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}$

On differentiating both sides w.r.t. $x$,

$f(x)=\cos\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}\times \{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}'$ [using chain rule]

$=\cos\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}\times \{{\dfrac{(x-\sin[\dfrac{x}{x-\sin x}])(x)'-x(x-\sin[\dfrac{x}{x-\sin x}])'}{(x-\sin[\dfrac{x}{x-\sin x}])^2}}\}$ [Using quotient rule]

$=\cos\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}\times \{{\dfrac{(x-\sin[\dfrac{x}{x-\sin x}])-x(1-\cos[\dfrac{x}{x-\sin x}])[\dfrac{x}{x-\sin x}]'}{(x-\sin[\dfrac{x}{x-\sin x}])^2}}\}$

$=\cos\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}\times \{{\dfrac{(x-\sin[\dfrac{x}{x-\sin x}])-x(1-\cos[\dfrac{x}{x-\sin x}])[\dfrac{(x-\sin x)-x(1-\cos x)}{(x-\sin x)^2}]}{(x-\sin[\dfrac{x}{x-\sin x}])^2}}\}$

$=\cos\{{\dfrac{x}{x-\sin[\dfrac{x}{x-\sin x}]}}\}\times \{{\dfrac{(x-\sin[\dfrac{x}{x-\sin x}])-x(1-\cos[\dfrac{x}{x-\sin x}])[\dfrac{x\cos x-\sin x}{(x-\sin x)^2}]}{(x-\sin[\dfrac{x}{x-\sin x}])^2}}\}$

This is our required $f'(x)$.