f-x-2sinx-2-cos-2-x-2-f-x-

Question Number 149291 by mathdanisur last updated on 04/Aug/21

f (x) = 2 {s i n x}^{2} - {c o s}^{2} x - 2

f^{^{'}} (x) = ?

Answered by nimnim last updated on 04/Aug/21

f^{'} (x) = 4 x . {c o s x}^{2} + 2 c o s x . s i n x - 0 = 4 x . c o s (x^{2}) + s i n (2 x)

Answered by Ar Brandon last updated on 04/Aug/21

f (x) = 2 \sin (x^{2}) - \cos^{2} x - 2

f^{'} (x) = 4 x \cos (x^{2}) + 2 \sin x \cos x

= 4 x \cos (x^{2}) + \sin 2 x

Commented by Ar Brandon last updated on 04/Aug/21

Haha! J^{'} avais commis cette m \hat{e m e} erreur

avant de m^{'} en rendre compte, apr \overset{`}{e s} avoir

vu les diff \overset{´}{e r e n t e s} solutions propos \overset{´}{e e s} .

Commented by Ar Brandon last updated on 04/Aug/21

Commented by mathdanisur last updated on 04/Aug/21

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