derivative of ln(h+1/(2ax^2+b))

解答

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

- \frac{2}{x ( 2 ah x ^{2} + 2 ab x ^{2} + 1 )}

求解步骤

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

将

a, b, h

视为常数

使用连锁律:

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

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

\frac{d}{d x} (h + \frac{1}{2 a x ^{2}} + b) = - \frac{1}{a x ^{3}}

= \frac{1}{h + \frac{1}{2 a x ^{2}} + b} (- \frac{1}{a x ^{3}})

化简

\frac{1}{h + \frac{1}{2 a x ^{2}} + b} (- \frac{1}{a x ^{3}}) : - \frac{2}{x ( 2 ah x ^{2} + 2 ab x ^{2} + 1 )}

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

x \to 0 lim (\frac{1 - cos ( 5 x )}{sin ( 3 x )})

\int_{- 8}^{8} x 64 - x^{2} d x

\int sin (4 t) sin (2 t) d t

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

s l o p e (9, 7), (6, - 2)