f^'(x)=-cos(x)

解答

f^{^{'}} (x) = - cos (x)

f (x) = - sin (x) + c_{1}

求解步骤

f^{'} (x) = - cos (x)

若

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

则

f (x) = \int g (x) d x

f (x) = \int - cos (x) d x

\int - cos (x) d x = - sin (x) + c_{1}

f (x) = - sin (x) + c_{1}

d er i v a t i v e y = 3 e^{x} + \frac{4}{3 x}

d er i v a t i v e \frac{3 x ^{3} + 7}{x}

\frac{d}{d x} ((3 x - 1)^{4})

\frac{d}{d x} (- cos (2 x) - sin (2 x) - \frac{1}{5} e^{x})

2 \frac{d y}{d x} + 6 y = 3