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Math Help - differentiation of tant + tsec^2t

  1. #1
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    differentiation of tant + tsec^2t

    Hello, I am getting confused as to how to differentiate this function:

    tant + tsec^2t

    I attempted it incorrectly as such:

    sec^2t + (1)(sec^2t) + (t)(2sint/cos^3t)
    by using product rule for the second part of the sum and using 2sinx/cos^3x as the differentiation of sec^2t

    Can someone please show me where I went wrong ?

    Thanks kindly
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  2. #2
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    Re: differentiation of tant + tsec^2t

    Quote Originally Posted by fran1942 View Post
    Hello, I am getting confused as to how to differentiate this function:

    tant + tsec^2t

    I attempted it incorrectly as such:

    sec^2t + (1)(sec^2t) + (t)(2sint/cos^3t)
    by using product rule for the second part of the sum and using 2sinx/cos^3x as the differentiation of sec^2t

    Can someone please show me where I went wrong ?

    Thanks kindly
    \displaystyle \begin{align*} \sec^2{t} = \left(\cos{t}\right)^{-2} \end{align*}, so if we let \displaystyle \begin{align*} y = \left(\cos{t}\right)^{-2} \end{align*}, we can let \displaystyle \begin{align*} u = \cos{t} \implies y = u^{-2} \end{align*}, then

    \displaystyle \begin{align*} \frac{du}{dt} &= -\sin{t} \\ \\ \frac{dy}{du} &= -2u^{-3} \\ &= -\frac{2}{\cos^3{t}} \\ \\ \frac{dy}{dt} &= \frac{2\sin{t}}{\cos^3{t}} \end{align*}

    I agree with your answer. What makes you think it's wrong?
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  3. #3
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    Re: differentiation of tant + tsec^2t

    2sint/cos^3t is also equal to 2sec^2ttant. Perhaps they have used this in their answer.
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  4. #4
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    Re: differentiation of tant + tsec^2t

    It is worth memorising that the derivative of secx is secxtanx. Can then use chain rule for other expressions
    Example For sec^2x (which means (secx)^2) we get 2secxsecxtanx=2sec^2xtanx
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