derivative of y=tan^2(sqrt(x))

解答

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

y = - x + 2 (x tan (x) + ln (cos (x))) + c_{1}

求解步骤

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

若

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

则

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

y = \int tan^{2} (x) d x

\int tan^{2} (x) d x = - x + 2 (x tan (x) + ln (cos (x))) + c_{1}

y = - x + 2 (x tan (x) + ln (cos (x))) + c_{1}

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

n = 1 \sum \infty \frac{5 n}{4 n + 15}

x \to - 1 - lim (\frac{x + 1}{x ^{4} - 1})

n = 0 \sum \infty \frac{1}{2 ^{n - 1}}

d er i v a t i v e \frac{x ^{4}}{2} - \frac{2}{x ^{4}} + 6 x