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Math Help - differentiating arctan

  1. #1
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    differentiating arctan

    Hi there,

    I came across a rule that says the derivative of arctanx is 1/(1 + x^2). How would you differentiatie arctanx to get this?

    Also, is there a way to show fractions with one factor on top of another, rather than side by side as in my example?

    Thanks,

    jjmclell
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  2. #2
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    y = \arctan x \implies \tan y = x \implies y'\sec ^2 y = 1.

    Can you take it from there?
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  3. #3
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    Let f(x) = \tan^{-1} x this function is differenciable on \mathbb{R}. Let g(x) = \tan x \mbox{ on }(-\pi/2,\pi/2) which is also differenciable. Then, g(f(x))=x. Using the chain rule f'(x)g'(f(x))=1 thus f'(x)\sec^2 (\tan^{-1} x)=1 thus f'(x)(1+x^2) = 1\implies f'(x) = \frac{1}{1+x^2}.
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  4. #4
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    Quote Originally Posted by jjmclell View Post
    Hi there,

    I came across a rule that says the derivative of arctanx is 1/(1 + x^2). How would you differentiatie arctanx to get this?

    Also, is there a way to show fractions with one factor on top of another, rather than side by side as in my example?

    Thanks,

    jjmclell
    Put u=\arctan(x) , then \tan(u)=x, so:

    \frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1

    and from this point its just algebra and trig identities.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Put u=\arctan(x) , then \tan(u)=x, so:

    \frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1

    and from this point its just algebra and trig identities.

    RonL

    Ok, I clearly see that if u=arctan(x) then tan(u)=x. However, why are we suddenly doing \frac{d}{dx}\tan(u)? Aren't we interested in \frac{du}{dx}?
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  6. #6
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    Quote Originally Posted by jjmclell View Post
    Ok, I clearly see that if u=arctan(x) then tan(u)=x. However, why are we suddenly doing \frac{d}{dx}\tan(u)? Aren't we interested in \frac{du}{dx}?
    Look at what you get when you do that differentiation (not difficult I have done it for you)

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Put u=\arctan(x) , then \tan(u)=x, so:

    \frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1

    and from this point its just algebra and trig identities.

    RonL

    Ok, I see that we wind up with \sec^2(u)\frac{du}{dx} which contains \frac{du}{dx} which we can isolate if we set it equal to 1. I even see how we get the 1 since \tan(u) = x and \frac{dx}{dx} = 1, but how does \frac{d}{dx}\tan(u) = sec^2\frac{du}{dx}? I know the derivitave of \tan(u) = sec^2(u) but that doesn't explain why it's multiplied by \frac{du}{dx} in this example. Sorry, I haven't taken calculus since highschool and now I'm trying to relearn it.
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  8. #8
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    Quote Originally Posted by jjmclell View Post
    Ok, I see that we wind up with \sec^2(u)\frac{du}{dx} which contains \frac{du}{dx} which we can isolate if we set it equal to 1. I even see how we get the 1 since \tan(u) = x and \frac{dx}{dx} = 1, but how does \frac{d}{dx}\tan(u) = sec^2\frac{du}{dx}? I know the derivitave of \tan(u) = sec^2(u) but that doesn't explain why it's multiplied by \frac{du}{dx} in this example. Sorry, I haven't taken calculus since highschool and now I'm trying to relearn it.
    Its the chain rule:

    \frac{d}{dx}f(u(x))=u'(x) ~f'(u(x))

    Now in your case f=\tan, so you
    have:

     <br />
\frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}<br />

    RonL
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  9. #9
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    Ahh, I see. So when you're taking a trig function of another trig function, say as in this case where we have \tan(\arctan(x)), is this always considered a composite function?
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  10. #10
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    Quote Originally Posted by jjmclell View Post
    Ahh, I see. So when you're taking a trig function of another trig function, say as in this case where we have \tan(\arctan(x)), is this always considered a composite function?
    Yes, if you looked at post #3 that was exactly what I was doing.
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  11. #11
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    Wow, it took me a while to figure out how \sec^2(u) = 1+x^2 but all is now clear. Thanks for everyone's help.
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