derivative of sqrt(tan(x+y))

解答

\frac{d}{d x} (tan (x) + y)

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

求解步骤

\frac{d}{d x} (tan (x) + y)

使用连锁律:

\frac{1}{2 tan ( x ) + y} \frac{d}{d x} (tan (x) + y)

= \frac{1}{2 tan ( x ) + y} \frac{d}{d x} (tan (x) + y)

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

= \frac{1}{2 tan ( x ) + y} (sec^{2} (x) + \frac{d y}{d x})

化简

\frac{1}{2 tan ( x ) + y} (sec^{2} (x) + \frac{d y}{d x}) : \frac{sec ^{2} ( x ) + \frac{d y}{d x}}{2 tan ( x ) + y}

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

6 = \frac{d x}{d t} + \frac{3 x}{20}

f (y) = 2 y^{2}

\int \frac{5 x ^{2} + 2 x + 2}{x ^{3} - 1} d x

d er i v a t i v e x ln (x)

t an g e n t f (x) = - 8 x^{2} - 7, at x = 1