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Finding the derivative of

  1. May 9, 2005 #1
    I am trying to do a question for calculus. I am supposed to find a minimum time for something.. I know how to do the question but I am stuck on how to find the derivative of this function it really confuses me.

    (sq. root 1^2+x^2)/3 + 3-x/5
    I need to find dt/dx... and I am clueless on how to find this???
    THANKS TONS!
     
  2. jcsd
  3. May 9, 2005 #2
    [tex]d\sqrt{Y}=\frac{dY}{2\sqrt{Y}}[/tex] Now for Y = A+B, the rule is:

    Y'=A'+B'. In the case of a form such as S/T, you can employ the rule:

    d(S/T) =[tex]\frac{S'T-ST'}{T^2}[/tex]

    However, in the problem above we don't need to use that since: d(X/3) is just X'/3. Again in the case

    [tex]\sqrt{\frac{1+x^2}{3}}=\frac{\sqrt{1+x^2}}{\sqrt3}[/tex] which is simpler to work with.
     
    Last edited: May 9, 2005
  4. May 10, 2005 #3

    James R

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    To differentiate

    [tex]t= \frac{\sqrt{1 + x^2}}{3}[/tex]

    put [itex]u=1 + x^2[/itex], then use the chain rule:

    [tex]\frac{dt}{dx} = \frac{dt}{du} \frac{du}{dx}[/tex]
     
    Last edited: May 10, 2005
  5. May 10, 2005 #4
    Or pulling out constants might make it seem simpler.

    [tex]\frac{1}{3}*\frac{d\sqrt{1+x^2}}{dx}[/tex] etc...
     
  6. May 10, 2005 #5

    HallsofIvy

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    Most people find it easier to think of "square root" as "1/2 power". That is,
    To find the derivative of [tex]\frac{\sqrt{1+x^2}}{3} + 3- \frac{x}{5}[/itex], write it as [tex]\frac{1}{3}(1+ x^2)^{\frac{1}{2}}+ 3- \frac{x}{5}[/itex]

    Then the derivative is [itex]\frac{1}{3}\frac{1}{2}(1+ x^2)^{-\frac{1}{2}}{2x}- \frac{1}{5}[/tex]
     
    Last edited: May 10, 2005
  7. May 10, 2005 #6
    You Guys Are Awesome Thanks Sooo Much!!!
     
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