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解答 - 导数

\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + 0 + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + 20 e n s x

\frac{d}{dx}[s]\times e\times n\times (10 x^{2})+0+s\times e\times \frac{d}{dx}[n]\times (10 x^{2})+20 e n s x

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逐步解答

1. 求导数

19 个额外步骤

扩展乘法的导数。

$\frac{d}{d x} [s \times e \times n \times (10 x^{2})] = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}]$

扩展乘法的导数。

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s \times e \times n \times (10 x^{2})] = \frac{d}{d x} [s \times (e \times n \times (10 x^{2}))]$

应用导数的乘法规则。

$\frac{d}{d x} [s \times (e \times n \times (10 x^{2}))] = \frac{d}{d x} [s] \times (e \times n \times (10 x^{2})) + s \times \frac{d}{d x} [e \times n \times (10 x^{2})]$

扩展乘法的导数。

$\frac{d}{d x} [s] \times (e \times n \times (10 x^{2})) + s \times \frac{d}{d x} [e \times n \times (10 x^{2})] = \frac{d}{d x} [s] \times (e \times n \times (10 x^{2})) + s (\frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}])$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [e \times n \times (10 x^{2})] = \frac{d}{d x} [e \times (n \times (10 x^{2}))]$

应用导数的乘法规则。

$\frac{d}{d x} [e \times (n \times (10 x^{2}))] = \frac{d}{d x} [e] \times (n \times (10 x^{2})) + e \times \frac{d}{d x} [n \times (10 x^{2})]$

扩展乘法的导数。

应用导数的乘法规则。

$\frac{d}{d x} [n \times (10 x^{2})] = \frac{d}{d x} [n] \times (10 x^{2}) + n \times \frac{d}{d x} [10 x^{2}]$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [e] \times (n \times (10 x^{2})) + e \times (\frac{d}{d x} [n] \times (10 x^{2}) + n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times (\frac{d}{d x} [n] \times (10 x^{2}) + n \times \frac{d}{d x} [10 x^{2}])$

可以通过分别乘以每个数，然后再加或减结果，来乘以两个数的和或差。

$\frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times (\frac{d}{d x} [n] \times (10 x^{2}) + n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times (\frac{d}{d x} [n] \times (10 x^{2})) + e \times (n \times \frac{d}{d x} [10 x^{2}]))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times (\frac{d}{d x} [n] \times (10 x^{2})) + e \times (n \times \frac{d}{d x} [10 x^{2}])) = \frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times (n \times \frac{d}{d x} [10 x^{2}]))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times (n \times \frac{d}{d x} [10 x^{2}])) = \frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}])$

加法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [e] \times n \times (10 x^{2}) + (e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}]$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times (e \times n \times (10 x^{2})) + s (\frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s (\frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}])$

可以通过分别乘以每个数，然后再加或减结果，来乘以两个数的和或差。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s (\frac{d}{d x} [e] \times n \times (10 x^{2}) + e \times \frac{d}{d x} [n] \times (10 x^{2}) + e \times n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times (\frac{d}{d x} [e] \times n \times (10 x^{2})) + s \times (e \times \frac{d}{d x} [n] \times (10 x^{2})) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}]))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times (\frac{d}{d x} [e] \times n \times (10 x^{2})) + s \times (e \times \frac{d}{d x} [n] \times (10 x^{2})) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}])) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times (e \times \frac{d}{d x} [n] \times (10 x^{2})) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}]))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times (e \times \frac{d}{d x} [n] \times (10 x^{2})) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}])) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}]))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times (e \times n \times \frac{d}{d x} [10 x^{2}])) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}])$

加法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + (s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}]$

常数的导数总是零。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times \frac{d}{d x} [e] \times n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}] = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}]$

应用导数的乘法规则。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times \frac{d}{d x} [10 x^{2}] = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (\frac{d}{d x} [10] \times x^{2} + 10 \times \frac{d}{d x} [x^{2}])$

常数的导数总是零。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (\frac{d}{d x} [10] \times x^{2} + 10 \times \frac{d}{d x} [x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (0 x^{2} + 10 \times \frac{d}{d x} [x^{2}])$

一个数乘以零总是等于零。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (0 x^{2} + 10 \times \frac{d}{d x} [x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (0 + 10 \times \frac{d}{d x} [x^{2}])$

一个数加零，其结果不变。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n (0 + 10 \times \frac{d}{d x} [x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times \frac{d}{d x} [x^{2}])$

计算x的n次幂的导数。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times \frac{d}{d x} [x^{2}]) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times (2 x^{2 - 1}))$

从一个数字中减去一。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times (2 x^{2 - 1})) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times (2 x^{1}))$

乘法可以以不同的方式分组，但结果仍然相同。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (10 \times (2 x^{1})) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times ((10 \times 2) \times x^{1})$

将两个整数倍乘在一起。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times ((10 \times 2) \times x^{1}) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x^{1})$

简化算术表达式。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x^{1}) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x)$

简化算术表达式。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + s \times 0 n \times (10 x^{2}) + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + 0 + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x)$

简化算术表达式。

$\frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + 0 + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + s \times e \times n \times (20 x) = \frac{d}{d x} [s] \times e \times n \times (10 x^{2}) + 0 + s \times e \times \frac{d}{d x} [n] \times (10 x^{2}) + 20 e n s x$

我们做得怎么样？

给我们反馈

为什么学习这个

是否想过如何预测未来？导数就是你的水晶球！

想象一下：你是一个想要捕捉最大浪的冲浪者。你如何知道它什么时候到来？导数可以告诉你什么时候它在最高点！

火箭科学：打算发射火箭去火星？导数告诉我们最佳的燃烧速率，以便最大限度地减少燃料消耗和增加距离！

股市：在股市交易吗？导数可以显示股票价格变化的速度，有助于预测最佳购买或销售时机。

动画：喜欢动画电影吗？艺术家们使用导数顺畅地改变角色的动作和表情，使他们感觉更像真实的生活。

工程学：正在设计一座桥或一座摩天大楼？导数可以帮助确定材料的应力和应变变化率，确保你的结构的安全。

简言之，导数就像是理解变化并在现实生活中做出预测的秘密代码。所以，我们一起来解开这个代码，成为我们未来的主人吧!

术语和主题

导数

解答 - 导数

逐步解答

1. 求导数

为什么学习这个

术语和主题

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