Calculus derivative Flashcards

(17 cards)

1
Q
  1. 𝑑/𝑑π‘₯ [𝑐𝑓(π‘₯)] =
A

𝑐𝑓′(π‘₯)

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2
Q
  1. 𝑑/𝑑π‘₯ [𝑓(π‘₯) Β± 𝑔(π‘₯)] =
A

𝑐𝑓′(π‘₯)

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3
Q
  1. Power Rule:𝑑/𝑑π‘₯ [π‘₯𝑛] = 𝑛π‘₯π‘›βˆ’1
A
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4
Q
  1. Product Rule:𝑑/𝑑π‘₯ [𝑓(π‘₯)𝑔(π‘₯)] = 𝑓(π‘₯)𝑔′(π‘₯) + 𝑔(π‘₯)𝑓′(π‘₯)
A
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5
Q
  1. Quotient Rule:𝑑/𝑑π‘₯ [𝑓(π‘₯)/𝑔(π‘₯)] =𝑔(π‘₯)𝑓′(π‘₯) βˆ’ 𝑓(π‘₯)𝑔′(π‘₯)/[𝑔(π‘₯)]2
A
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6
Q
  1. Chain Rule:𝑑/𝑑π‘₯ [𝑓(𝑔(π‘₯))] = 𝑓′(𝑔(π‘₯)) βˆ™ 𝑔′(π‘₯)
A
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7
Q
  1. Inverse Rule:𝑑/𝑑π‘₯ [𝑔(π‘₯)] =1/𝑓′(𝑔(π‘₯))
A
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8
Q
  1. 𝑑/𝑑π‘₯ [𝑒π‘₯] = 𝑒π‘₯
A
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9
Q
  1. 𝑑/𝑑π‘₯ [π‘Žπ‘₯] = π‘Žπ‘₯ ln π‘Ž
A
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10
Q
  1. 𝑑/𝑑π‘₯ [ln π‘₯] =1/π‘₯
A
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11
Q
  1. 𝑑/𝑑π‘₯ [logπ‘Ž π‘₯] =1/π‘₯ ln π‘Ž
A
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12
Q
  1. 𝑑/𝑑π‘₯ [sin π‘₯] = cos π‘₯
A
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13
Q
  1. 𝑑/𝑑π‘₯ [cos π‘₯] = βˆ’ sin π‘₯
A
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14
Q
  1. 𝑑/𝑑π‘₯ [tan π‘₯] = sec2 π‘₯
A
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15
Q
  1. 𝑑/𝑑π‘₯ [cot π‘₯] = βˆ’ csc2 π‘₯
A
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16
Q
  1. 𝑑/𝑑π‘₯ [sec π‘₯] = sec π‘₯ tan π‘₯
17
Q
  1. 𝑑/𝑑π‘₯ [csc π‘₯] = βˆ’ csc π‘₯ cot π‘₯