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michaela

  • 5 years ago

find the derivative of y=sin^3 x

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  1. anonymous
    • 5 years ago
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    What you want to remember here is the chain rule. Take each part by itself. =sin^3(x) = 3sin^2(x) = 3cos^2(x)*1 = 3cos^2(x) Remember the 1 from the inner x, if it's something else then x it could change the result. Unless I'm forgetting the chain rule myself, but it should be correct. :)

  2. michaela
    • 5 years ago
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    THANK YOU :)

  3. anonymous
    • 5 years ago
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    You're welcome :)

  4. apples
    • 5 years ago
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    This is incorrect. The inner function is \[f(x) = \sin{x}\] and the outer function is \[g(x) = x^{3}\] Thus \[\frac{d}{dx} \sin^3{x} = \frac{d}{dx} g \circ f = \frac{dg}{dx}(f(x)) \cdot \frac{df}{dx} = 3\sin^2{x} \cdot \cos{x} \]

  5. anonymous
    • 5 years ago
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    Yeah, thought I did something wrong. Good catch!

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