How do I find the derivative of this (t^6 +1)^99?

In the parenthesis do you add the 99 and 6 or multiply? so would it be 594^5 or 105t^5? Or do you even distribute the 99?? Thanks

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  • 8 years ago
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    Use the Differentiation Chain Rule and the Differentiation Power Rule.

    The derivative is

    [ 99 * (t^6 + 1)^98 ] * [ (d/dt) (t^6 + 1) ]

    = [ 99 * (t^6 + 1)^98 ] * 6t^5

    = 99 * 6t^5 * (t^6 + 1)^98

    = 594t^5 * (t^6 + 1)^98

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    That's the simplest way to write it, I think. Unless you WANT to use the binomial theorem and expand and write down all the terms of (t^6 +1)^98 . . . ;)

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