derivative f(x)=-4x^3-cos(x)+2x

الحلّ

derivative of

f (x) = - 4 x^{3} - cos (x) + 2 x

- 12 x^{2} + sin (x) + 2

خطوات الحلّ

\frac{d}{d x} (- 4 x^{3} - cos (x) + 2 x)

(f \pm g)^{'} = f^{'} \pm g^{'}

:استعمل قانون الجمع

= - \frac{d}{d x} (4 x^{3}) - \frac{d}{d x} (cos (x)) + \frac{d}{d x} (2 x)

\frac{d}{d x} (4 x^{3}) = 12 x^{2}

\frac{d}{d x} (cos (x)) = - sin (x)

\frac{d}{d x} (2 x) = 2

= - 12 x^{2} - (- sin (x)) + 2

بسّط

= - 12 x^{2} + sin (x) + 2

y = x^{3}

e^{x} sin (x)

(- 4, \frac{3 π}{4})

f (x) = \frac{1}{x}

y = - \frac{1}{2} x + 3