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Math Help - Differentiate the following function

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    Differentiate the following function

    f(t)=5/7(t^7+8)
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by plstevens View Post
    f(t)=5/7(t^7+8)

    I'm going to assume the equation is:

    f(t)=\frac{5}{7(t^7+8)}

    If so, then you have two options.

    As the problem is, you can use the quotient rule: \left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}

    Or you can bring the denomintor up and give it a negative exponent, giving us the following equation.

    f(t)=\frac{5}{7}(t^7+8)^{-1}

    I prefer to solve this because it uses the chain rule and I think the chain rule is easier to work with than the quotient rule.

    f'(t)=\frac{5}{7}(-1)(t^7+8)^{-2}(7t^6)=-\frac{5t^6}{(t^7+8)^2}
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    its (5/7)*(t^7+8)
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    Quote Originally Posted by plstevens View Post
    its (5/7)*(t^7+8)
    Expand and use the power rule.
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    7*(5/7)t^7-1+(40/7)=(35/7)t^6+40/7=5t^6+40/7
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    did i do that right,its coming up wrong, can u show me how to do the power rule
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by plstevens View Post
    did i do that right,its coming up wrong, can u show me how to do the power rule
    f(t) = \frac{5}{7}(t^7 + 8)

    f(t) = \frac{5}{7}t^7 + \frac{5}{7} \cdot 8

    Thus
    f^{\prime}(t) = \frac{5}{7} \cdot 7 \cdot t^{7 - 1}

    f^{\prime}(t) = 5t^6

    The derivative of the function
    f(x) = ax^n

    is
    f^{\prime}(x) = a \cdot n \cdot x^{n - 1}
    as has been stated a couple of times in other threads. What part of this formula is causing you so much trouble?

    -Dan
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