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Idealist

  • 3 years ago

Find the second derivative of f(x)=sinxcosx.

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  1. inkyvoyd
    • 3 years ago
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    Well, what's the first derivative?

  2. inkyvoyd
    • 3 years ago
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    use the product rule...

  3. saifoo.khan
    • 3 years ago
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  4. Idealist
    • 3 years ago
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    The first derivative is (cosx)^2-(sinx)^2. But how do I take the derivative of that?

  5. inkyvoyd
    • 3 years ago
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    well, remember that cos(a+b)=cos a cos b-sin a sin b...

  6. hcmathkim
    • 3 years ago
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    use the chain rule

  7. inkyvoyd
    • 3 years ago
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    in other words, use the double angle formula for to reduce the following expression you have obtained. And then use the chain rule.

  8. Sujay
    • 3 years ago
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    \[f'(x)=cosx(\cos(x))+(-\sin(x)\sin(x))\] Product Rule and Derivatives of Trigs Simplify. \[f'(x)=\cos ^{2}x-\sin ^{2}x\] \[f''(x)=2\cos(x)(-\sin(x))-2\sin(x)(\cos(x))\] Derivative of Trig and Chain Rule

  9. hcmathkim
    • 3 years ago
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    bring the power down, keep the inside, differentiate inside

  10. inkyvoyd
    • 3 years ago
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    Since no one seems to get it USE COS^2 X-SIN^2 X=COS(2x) TO SIMPLIFY EXPRESSION

  11. inkyvoyd
    • 3 years ago
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    simplified expression is dy/dx=cos(2x) then take the dervative of that

  12. inkyvoyd
    • 3 years ago
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    @Idealist are you followign?

  13. Idealist
    • 3 years ago
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    Yes. I got it. Thanks for the help, guys.

  14. inkyvoyd
    • 3 years ago
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    I like medals. LOL but yeah, nice to see :) let me know your final answer so I can check it

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