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anonymous

  • 5 years ago

Help please! can someone show me the steps of how to differentiate the following with respect to x 3sin^2x+sec2x

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  1. anonymous
    • 5 years ago
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    ddxsin(ax)dx=acos(ax) use this along with chain rule for 3sin^2x

  2. anonymous
    • 5 years ago
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    differentiate each term in turn for 3 sin^2x use the chain rule f'(x) = f'(g(x) * g'(x) also use chain rule for second term

  3. anonymous
    • 5 years ago
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    d/dx [3sin^2(x)] = 6sin x cos x

  4. anonymous
    • 5 years ago
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    yup

  5. anonymous
    • 5 years ago
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    Let y=sec 2x dy/dx=d{sec 2x}/dx =sec 2x tan2x×d(2x)/dx USing chain rule =sec 2x tan2x×(2) =2sec 2x tan2x

  6. anonymous
    • 5 years ago
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    @jetly did you use the chain rule to get 6sin x cos x?

  7. anonymous
    • 5 years ago
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    yes kinzan.

  8. anonymous
    • 5 years ago
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    are you happy with the f'(x) notation or do you usually use the Leibnitz fom r kinsan : - dy/dx etc

  9. anonymous
    • 5 years ago
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    *Leibnitz form

  10. anonymous
    • 5 years ago
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    i use the the f'(x) notation but do not get how jetly go 2sec 2x tan2x as his final answer... arnt you meant to use a constant 'u' in the chanirule?

  11. anonymous
    • 5 years ago
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    you can use 'u' for the 'inside function' when using the chain rule - some people find it easier to understand

  12. anonymous
    • 5 years ago
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    i'll differentiate for you using 'u' and leibnitz notation; let y = sec2x and let u = 2x so y = secu dy/du = secu tanu du/dx = 2 so dy/dx = dy/du * du/dx = 2 secu tan u = 2 sec2x tan 2x

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