derivative of 1/(4(1+x^{3/4)})

解答

\frac{d}{d x} (\frac{1}{4 ( 1 + x ) ^{\frac{3}{4}}})

- \frac{3}{16 ( 1 + x ) ^{\frac{7}{4}}}

求解步骤

\frac{d}{d x} (\frac{1}{4 ( 1 + x ) ^{\frac{3}{4}}})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d x} (\frac{1}{( 1 + x ) ^{\frac{3}{4}}})

使用指数法则:

\frac{1}{a} = a^{- 1}

= \frac{1}{4} \frac{d}{d x} ((1 + x)^{- \frac{3}{4}})

使用连锁律:

- \frac{3}{4 ( 1 + x ) ^{\frac{7}{4}}} \frac{d}{d x} (1 + x)

= - \frac{3}{4 ( 1 + x ) ^{\frac{7}{4}}} \frac{d}{d x} (1 + x)

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

= \frac{1}{4} (- \frac{3}{4 ( 1 + x ) ^{\frac{7}{4}}} \cdot 1)

化简

\frac{1}{4} (- \frac{3}{4 ( 1 + x ) ^{\frac{7}{4}}} \cdot 1) : - \frac{3}{16 ( 1 + x ) ^{\frac{7}{4}}}

= - \frac{3}{16 ( 1 + x ) ^{\frac{7}{4}}}

\int \frac{sin ( x )}{cos ^{5} ( x )} d x

\frac{d}{d x} (5 x^{2} - 1)

\frac{d N}{d t} + N = Nt e^{(t + 2)}

n = 1 \sum \infty \frac{n !}{2 n - 1}

y^{^{'}} = \frac{5}{y + 2}