derivative 1/(5t^2-6t+2)

解答

导数

\frac{1}{5 t ^{2} - 6 t + 2}

- \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

求解步骤

\frac{d}{d t} (\frac{1}{5 t ^{2} - 6 t + 2})

使用指数法则:

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

= \frac{d}{d t} ((5 t^{2} - 6 t + 2)^{- 1})

使用连锁律:

- \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} \frac{d}{d t} (5 t^{2} - 6 t + 2)

= - \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} \frac{d}{d t} (5 t^{2} - 6 t + 2)

\frac{d}{d t} (5 t^{2} - 6 t + 2) = 10 t - 6

= - \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} (10 t - 6)

化简

- \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} (10 t - 6) : - \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

= - \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

\int \frac{y}{y + 1} d y

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