derivative of tan(x-x+1)

解答

\frac{d}{d x} (tan (x) - x + 1)

tan^{2} (x)

求解步骤

\frac{d}{d x} (tan (x) - x + 1)

使用微分加减法定则:

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

= \frac{d}{d x} (tan (x)) - \frac{d x}{d x} + \frac{d}{d x} (1)

\frac{d}{d x} (tan (x)) = sec^{2} (x)

\frac{d x}{d x} = 1

\frac{d}{d x} (1) = 0

= sec^{2} (x) - 1 + 0

化简

sec^{2} (x) - 1 + 0 : tan^{2} (x)

= tan^{2} (x)

\frac{d}{d x} (1 + x^{256} - 1)

d er i v a t i v e f (x) = 19 x e^{x}

\int sin (x) e^{3 x} d x

\int x^{2} cos (\frac{1}{9} x) d x

\frac{d}{d x} (sec (x) tan^{2} (x))