derivative f(x)=ln(x+sqrt(x^2+1))

解答

导数

f (x) = ln (x + x^{2} + 1)

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

求解步骤

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

使用连锁律:

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

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

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

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

化简

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

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

\int (7 + e) d x

\int \frac{( x ^{2} )}{x ^{2} + 64} d x

\frac{\partial}{\partial x} (- 3 sin (3 x))

t an g e n t 4 x - \frac{x ^{3}}{4}

\int - y sin (y) d y