f-x-x-sinx-0-lt-x-lt-pi-2-find-f-x-

Question Number 69680 by 20190927 last updated on 26/Sep/19

f (x) = x^{sinx}, 0 < x < \frac{π}{2} find f^{'} (x)

Answered by MJS last updated on 26/Sep/19

\frac{d}{d x} [u^{v}] = u^{v} (\frac{u^{'} v}{u} + v^{'} \ln u)

\Rightarrow

f^{'} (x) = x^{\sin x} (\frac{\sin x}{x} + \cos x \ln x)

Commented by 20190927 last updated on 26/Sep/19

thank you so much

Answered by Henri Boucatchou last updated on 26/Sep/19

f^{'} (x) = {(e^{s i n x l n x})}^{'} = {(s i n x l n x)}^{'} f (x)

= (c o s x l n x + \frac{s i n x}{x}) x^{s i n x}