derivative of sqrt(x+\sqrt{x+1)}

解答

\frac{d}{d x} (x + x + 1)

\frac{2 x + 1 + 1}{4 x + 1 x + x + 1}

求解步骤

\frac{d}{d x} (x + x + 1)

使用连锁律:

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

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

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

= \frac{1}{2 x + x + 1} (1 + \frac{1}{2 x + 1})

化简

\frac{1}{2 x + x + 1} (1 + \frac{1}{2 x + 1}) : \frac{2 x + 1 + 1}{4 x + 1 x + x + 1}

= \frac{2 x + 1 + 1}{4 x + 1 x + x + 1}

a re a y = - x^{2} + 16, y = \frac{- x}{4} - 1, 4 (y + 6.5) = 6.5 x

d er i v a t i v e (3 x^{2} - 6 x + 6) e^{- 5 x}

\int_{0}^{2} sin^{2} (x co) s^{2} 2 x

\int \frac{5}{4 + 9 x ^{2}} d x

\frac{\partial}{\partial y} (x^{2} + 6 x y + y^{4})