derivative of log_{10}(x^2+x+1)

解答

\frac{d}{d x} (lo g_{10} (x^{2} + x + 1))

\frac{2 x + 1}{ln ( 10 ) ( x ^{2} + x + 1 )}

求解步骤

\frac{d}{d x} (lo g_{10} (x^{2} + x + 1))

使用对数计算法则:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( x ^{2} + x + 1 )}{ln ( 10 )})

将常数提出:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (x^{2} + x + 1))

使用连锁律:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{x ^{2} + x + 1} (2 x + 1)

化简

\frac{1}{ln ( 10 )} \cdot \frac{1}{x ^{2} + x + 1} (2 x + 1) : \frac{2 x + 1}{ln ( 10 ) ( x ^{2} + x + 1 )}

= \frac{2 x + 1}{ln ( 10 ) ( x ^{2} + x + 1 )}

t an g e n t f (x) = \frac{2 x + 3}{3 x - 2}

a re a 3 ln (x), x ln (x), 1, 2

\int sec^{2} (3 x) tan^{3} (3 x) d x

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