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

解答

导数

f (x) = ln (x + 1)

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

求解步骤

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

使用连锁律:

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

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

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

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

化简

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

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