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Math Help - Differential of tan

  1. #1
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    Differential of tan

    Does anyone know how I get from
    y=2 tan-1 (et)
    to
    dy/dx = sech t ?
    (the minus one after tan and the t following the e are powers!!)
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    Quote Originally Posted by bbbabybel View Post
    Does anyone know how I get from
    y=2 tan-1 (et)
    to
    dy/dx = sech t ?
    (the minus one after tan and the t following the e are powers!!)
    dy/dx = 0 since there's no x in the expression.

    To find dy/dt, use the chain rule:

    \frac{dy}{dt} = \frac{2 e^t}{1 + (e^t)^2} = \frac{2 e^t}{1 + e^{2t}}.

    Now divide numertaor and denominator by e^t:

    \frac{dy}{dt} = \frac{2}{e^{-t} + e^{t}} = \text{sech}\, t by definition.
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  3. #3
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    Quote Originally Posted by bbbabybel View Post
    Does anyone know how I get from
    y=2 tan-1 (et)
    to
    dy/dx = sech t ?
    (the minus one after tan and the t following the e are powers!!)
    y=2arctan(e^t)

    solve for e^t (not really necessary, but I figure you'll see it better this way)

    tan(y/2)=e^t

    take the derivative

    \frac 12 sec^2(y/2) ~y\prime=e^t

    solve for y'

    y\prime=2~cos^2(y/2) ~e^t

    substitute for y from the first equation

    y\prime=2~cos^2(arctan(e^t)) ~e^t

    simplify (draw a triangle if necessary... or just memorize the derivative of arctangent and save yourself this extra work

    y\prime=\frac{2e^t}{e^{2t}+1}

    divide numerator and denominator by e^t

    y\prime=\frac{2}{e^t+\frac{1}{e^t}}

    y\prime=\frac{2}{e^t+e^{-t}}

    an equation is equal to 1 over it's reciprocal

    y\prime=\frac {~~~~~1~~~~~}{\frac{e^t+e^{-t}}{2}}

    the denominator is the equation for cosh(t)

    y\prime=\frac {1}{cosh(t)}

    and simplify

    y\prime=sech(t)
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    Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!
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    Or the chain rule rather....oh dear,I'm trying!
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  6. #6
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    Quote Originally Posted by bbbabybel View Post
    Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!
    Hello,

    (\tan^{-1} t)'=\frac{1}{1+t^2}


    Thus, using the chain rule :

    (\tan^{-1} f(x))'=\frac{f'(x)}{1+f^2(x)}
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  7. #7
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    While we're on the lovely subject, what's the differential of sech x?
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  8. #8
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    Quote Originally Posted by bbbabybel View Post
    While we're on the lovely subject, what's the differential of sech x?


    sech(x) = 1 / cosh(x)

    We know that the derivative of cosh(x) is sinh(x).

    The derivative of something like 1/f(x) is -f'(x)/f(x)

    ---> the derivative of sech(x) is \frac{-\sinh(x)}{\cosh^2(x)}=\frac{-1}{\cosh(x)} \cdot \tanh(x)


    You can take a look at these explanations, here : Hyperbolic Secant -- from Wolfram MathWorld
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  9. #9
    Super Member angel.white's Avatar
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    Quote Originally Posted by bbbabybel View Post
    Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!
    Using the same process I used before

    y=arctan(x)

    tan(y) = x

    differentiate

    sec^2(y) y\prime = 1

    y\prime = cos^2(y)

    substitute from the first equation

    y\prime = cos^2(arctan(x))

    y\prime = \left(\frac 1{\sqrt{x^2+1}}\right)^2

    y\prime = \frac 1{x^2+1}

    and because y = arctan(x)

    \frac d{dx} ~arctan(x) = \frac 1{x^2+1}
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