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Calculus 1

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Derivative of 7^(SinX)... something to do with log/ln?

  • Calculus 1 -

    You have to use the chain rule.

    ln(7) * 7^(sin x) * cos (x)

  • Calculus 1 -

    let y = 7^(sinx)
    take ln of both sides
    ln y = ln (7^sinx)
    ln y = sinx (ln7)
    now use the rule for ln
    y' /y = ln7(cosx)
    y' = ln7(cosx) * (7^sinx)

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