(d^2)/(dx^2)(5cos(2x+10))

解答

\frac{d ^{2}}{d x ^{2}} (5 cos (2 x + 10))

- 20 cos (2 x + 10)

求解步骤

\frac{d ^{2}}{d x ^{2}} (5 cos (2 x + 10))

\frac{d}{d x} (5 cos (2 x + 10)) = - 10 sin (2 x + 10)

= \frac{d}{d x} (- 10 sin (2 x + 10))

\frac{d}{d x} (- 10 sin (2 x + 10)) = - 20 cos (2 x + 10)

= - 20 cos (2 x + 10)

q \to 18 lim (0.9 q^{2} - 6 q + 80)

d er i v a t i v e sec (θ) (θ - tan (θ))

f (x) = x^{2} - 10 x + 34 + x^{2} + 16

y^{^{'}} - 5 y = e^{- 2 x} y^{- 2}

\frac{d}{d x} (f (x)^{2} (x))