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mathcalculus

  • one year ago

H-E-L-P : HOW TO FIND DERIVATIVE SQUARE ROOT OF 8-6*X^2

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  1. mathcalculus
    • one year ago
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    \[\sqrt{8-6*X^2}\]

  2. satellite73
    • one year ago
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    the derivative of \(\sqrt{f(x)}\) is \[\frac{f'(x)}{2\sqrt{f(x)}}\]

  3. mathcalculus
    • one year ago
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    can we do this step by step please.

  4. satellite73
    • one year ago
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    the steps are what i wrote above take the derivative of \(8-6x^2\) and by the power rule, you get \(-12x\) that goes in the numerator in the denominator you put \(2\sqrt{8-6x^2}\)

  5. mathcalculus
    • one year ago
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    oh... why do we put that as the denominator?

  6. satellite73
    • one year ago
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    all that is left to do is cancel a 2 top and bottom

  7. satellite73
    • one year ago
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    the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) therefore by the chain rule the derivative of \(\frac{1}{f}\) is \(\frac{f'}{2\sqrt{f}}\) where \(f\) is any function

  8. satellite73
    • one year ago
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    for example, the derivative of \(\sqrt{\sin(x)}\) is \[\frac{\cos(x)}{2\sqrt{\sin(x)}}\]

  9. satellite73
    • one year ago
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    the derivative of \[\sqrt{3x^2+2x}\] is \[\frac{6x+2}{2\sqrt{3x^2+2x}}\]

  10. mathcalculus
    • one year ago
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    after i have -12x/ 2radical 8-6x^2.. how can i find the critical points?

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