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Math Help - find the differential of each function

  1. #1
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    find the differential of each function

    The question is find the differential of each function: a) y=(t+tant)^5 and b) y=√(z+1/z)
    Please Help Me. I only got the first steps:
    a) dy/dt = 5(t+tan t)(1 +sec^2 x)
    b) dy/dz= 0.5*(1/(z + 1/z)) * (1 - 1/z^2)
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  2. #2
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    Quote Originally Posted by watchthesky30 View Post
    The question is find the differential of each function: a) y=(t+tant)^5 and b) y=√(z+1/z)
    Please Help Me. I only got the first steps:
    a) dy/dt = 5(t+tan t)(1 +sec^2 x)
    b) dy/dz= 0.5*(1/(z + 1/z)) * (1 - 1/z^2)
    I think you mean "find the derivative" of each function.

    You will need to use the chain rule in each case.


    a) y = (t + \tan{t})^5.

    Let u = t + \tan{t} so that y = u^5.


    \frac{du}{dt} = 1 + \sec^2{t} = 1 + \tan^2{t} + 1 = 2 + \tan^2{t}.


    \frac{dy}{du} = 5u^4 = 5(t + \tan{t})^4.


    So \frac{dy}{dt} = 5(2 + \tan{t})(t + \tan{t})^4.



    b) y = \sqrt{z + \frac{1}{z}} = \left(z + z^{-1}\right)^{\frac{1}{2}}.


    Let u = z + z^{-1} so that y = u^{\frac{1}{2}}.


    \frac{du}{dz} = 1 - z^{-2} = 1 - \frac{1}{z^2} = \frac{z^2 - 1}{z^2}.


    \frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{z + \frac{1}{z}}}.


    Therefore \frac{dy}{dz} = \frac{z^2 - 1}{2z^2\sqrt{z + \frac{1}{z}}}.
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