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Math Help - derivatives using the quotient rule

  1. #1
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    derivatives using the quotient rule

    Find the derivative of the function below.


    a little help por favor
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  2. #2
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    Hello, mer1988!

    Differentiate: . z(t)\:=\:\frac{t^2+5t+2}{t+3}
    Quotient Rule . Given: y \:=\:\frac{f}{g}
    . . . . . . . . . . . .then: . y' \:=\:\frac{gf' - fg'}{g^2}


    We have: . z(t) \;=\;\frac{\overbrace{t^2+5t+2}^{f}}{\underbrace{\  ,t+3\,}_{g}}

    Then: . z'(t) \;=\;\frac{\overbrace{(t+3)}^{g}\overbrace{(2t+5)}  ^{f'} - \overbrace{(t^2+5t+2)}^f\overbrace{(1)}^{g'}}{\und  erbrace{(t+3)^2}_{g^2}} \;= \;\frac{2t^2+5t+6t+15 - t^2-5t-2}{(t+3)^2}

    Therefore: . z'(t)\;=\;\frac{t^2+6t+13}{(t+3)^2}

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  3. #3
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    Quote Originally Posted by mer1988 View Post
    Si no quieres usar la regla del cuociente, también podemos hacer lo siguiente

    \begin{aligned}<br />
z(t) &= \frac{{t^2 + 5t + 2}}<br />
{{t + 3}}\\<br />
&= \frac{{t(t + 3) + 2(t + 3) - 4}}<br />
{{t + 3}}\\<br />
&= t + 2 - 4(t + 3)^{ - 1}\\<br />
z'(t) &= 1 + \frac{4}<br />
{{(t + 3)^2 }}\\<br />
&= \frac{{t^2 + 6t + 13}}<br />
{{(t + 3)^2 }}<br />
\end{aligned}

    Saludos
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  4. #4
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    Find the derivative of the function below.


    hows about this one...?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mer1988 View Post
    Find the derivative of the function below.


    hows about this one...?
    why don't you try this one so we can see if you understand?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    why don't you try this one so we can see if you understand?
    ok, I came up with...

    (t^(-1/2)-t(^1/2)*t)/(t^2+9)
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