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unkabogable

how to get for the derivative of cos^2 (x)- sin^2 (x)?

  • one year ago
  • one year ago

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  1. agentx5
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    a la Chain Rule! :-D Do you know it?

    • one year ago
  2. Spacelimbus
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    If you want you can always replace cos^2(x) with 1-sin^2(x), if that's easier for you to derive.

    • one year ago
  3. unkabogable
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    i didn't know that..

    • one year ago
  4. Spacelimbus
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    then you have 1-2sin^2(x) for your new problem

    • one year ago
  5. Spacelimbus
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    it's still what @agentx5 suggested though,just an algebraic manipulation.

    • one year ago
  6. nitz
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    2cosx(-sinx)-2sinx(cosx) -2cosxsinx-2sinxcosx -4sinxcosx

    • one year ago
  7. agentx5
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    Tada! \(\large \frac{d}{dx}(\cos^2(x)-\sin^2(x))\) \(\large \frac{d}{dx}(\cos^2(x))-\frac{d}{dx}(\sin^2(x))\) \(\large 2 cos(x) (\frac{d}{dx}(cos(x)))-2 sin(x) (\frac{d}{dx}(sin(x)))\) \(\large -2 \cos(x) \sin(x) -2 \sin(x) \cos(x)\) \(\large -4 \sin(x) \cos(x)\)

    • one year ago
  8. agentx5
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    See how the \(\frac{d}{dx}\) is carried from the outer function into the inner? \(\large \frac{d}{dx} (x^3)^2 = 2 (x^3)^{2-1} \cdot \frac{d}{dx} x^3 = 2 (x^3)^{1} \cdot 3x^{3-1} \frac{d}{dx} x = 6x^5 \ \ \ \ dx\)

    • one year ago
  9. unkabogable
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    and...voila!! thanks! I'm a fan!!

    • one year ago
  10. agentx5
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    It's the same exact thing, just with sine & cosine instead of exponents or whatever. Nested functions within functions, chain rule. Make sense?

    • one year ago
  11. agentx5
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    In my mind, I treat the derivative as an operand, like a special form of multiplication/division or what-have-you

    • one year ago
  12. agentx5
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    Granted that's not TECHNICALLY what it is, but you can "distribute" across added/subtracted terms and it applies into nested functions with the chain rule... Chain Rule: for: \[u^n \] then: \[(n*u^{n-1})*\frac{d}{du}u\] Do the power rule, then take the derivative of what's inside the "shell"/"layer"/"whatever"

    • one year ago
  13. agentx5
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    Learn it well @unkabogable ! It's kind of critical in all Calculus levels (and it only get more complex).

    • one year ago
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