derivative of xcos(x^2+1-ln(3-x))

解答

\frac{d}{d x} (x cos (x^{2} + 1) - ln (3 - x))

cos (x^{2} + 1) - 2 x^{2} sin (x^{2} + 1) + \frac{1}{3 - x}

求解步骤

\frac{d}{d x} (x cos (x^{2} + 1) - ln (3 - x))

使用微分加减法定则:

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

= \frac{d}{d x} (x cos (x^{2} + 1)) - \frac{d}{d x} (ln (3 - x))

\frac{d}{d x} (x cos (x^{2} + 1)) = cos (x^{2} + 1) - 2 x^{2} sin (x^{2} + 1)

\frac{d}{d x} (ln (3 - x)) = - \frac{1}{3 - x}

= cos (x^{2} + 1) - 2 x^{2} sin (x^{2} + 1) - (- \frac{1}{3 - x})

化简

= cos (x^{2} + 1) - 2 x^{2} sin (x^{2} + 1) + \frac{1}{3 - x}

\int \frac{2 arctan ( n )}{1 + n ^{2}} d n

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 20} d x

\int \frac{x ^{3}}{x ^{2} + 2} d x

\int \frac{cos ( a x )}{1 - sin ( a x )} d x

n = 1 \sum \infty \frac{x ^{n}}{4 n - 1}