derivative f(x)=2x-xln|x|

解答

导数

f (x) = 2 x - x ln ∣ x ∣

- ln ∣ x ∣ + 1

求解步骤

\frac{d}{d x} (2 x - x ln ∣ x ∣)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (2 x) - \frac{d}{d x} (x ln ∣ x ∣)

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

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

= 2 - (ln ∣ x ∣ + 1)

化简

= - ln ∣ x ∣ + 1

n = 0 \sum \infty \frac{1 ^{n}}{2 n + 1}

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