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Derivation of tanx.

  1. Oct 28, 2004 #1
    [itex]f(x) = \tan(x) = \frac{\sin(x)}{\cos(x)}[/itex]
    Using: [itex](\Delta x = h)[/itex]

    [itex]f'(x) = \frac{f(x + h) - f(x)}{h}[/itex]

    [itex]f'(x) = \frac{\frac{\sin(x + h)}{\cos(x + h)} - \frac{sin(x)}{\cos(x)}}{h}[/itex]

    Where do I go from here?
     
    Last edited: Oct 28, 2004
  2. jcsd
  3. Oct 28, 2004 #2
    [itex]\sin(x + h) = -\sin(x)[/itex]
    [itex]\cos(x + h) = -cos(x)[/itex]

    [itex]f'(x) = \frac{\frac{-\sin(x)}{-\cos(x)} - \frac{\sin(x)}{\cos(x)}}{h}[/itex]

    Is this the correct step?
    hmpf, probably not..
     
    Last edited: Oct 28, 2004
  4. Oct 28, 2004 #3
    Maybe:

    [itex]sin(u + v) = \sin(u)\cos(v) + \cos(u)\sin(v)[/itex]

    Thus:

    [itex]f'(x) = \frac{\frac{\sin(x)\cos(h) + \cos(x)\sin(h)}{\cos(x)\cos(h) - \sin(x)\sin(h)} - \frac{sin(x)}{\cos(x)}}{h}[/itex]

    ?
     
  5. Oct 28, 2004 #4

    Galileo

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    First try finding the derivative of sin(x) and cos(x) first. (You probably know them already).
    Then use the quotient rule.
    [tex]\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}[/tex]
     
  6. Oct 28, 2004 #5
    Yes, I have no problems doing it that way.
    But, how about using the general formula for derivatives:

    [itex]f'(x) = \frac{f(x + h) - f(x)}{h}[/itex]

    Possible?
     
  7. Oct 28, 2004 #6
    definitely,

    Needs a bit of trigonometry result .....
    tan(A-B) = [tan(A) - tan(B)]/[1+tan(A)tan(B)]
    therefore,
    tan(A) - tan(B) = tan(A-B)*(1+tan(A)tan(B))

    tan(x+h)-tan(x) = tan(h)*(1+tan(x+h)tan(x))
    Can u finish off now?

    -- AI
     
  8. Oct 28, 2004 #7
    Hm..
    [itex]f'(x) = \frac{\tan(h) * (1 + \tan(x+h)\tan(x))}{h}[/itex]

    I'm not sure how I would proceed from here?
     
  9. Oct 28, 2004 #8
    i am sure u know,
    [tex]f'(x) = \lim_{h->0} \frac{f(x+h)-f(x)}{h}[/tex]

    so what is lim_{h->0} tan(h)/h ??

    -- AI
     
  10. Oct 28, 2004 #9
    [itex]\frac{tan(x)}{x} = \frac{\frac{\sin(x)}{\cos(x)}}{x} = \frac{\sin(x)}{x}\frac{1}{cos(x)} = 1[/itex]

    I'm not seeing the bigger picture though :)

    Hm.
    [itex]f'(x) = \frac{\tan(h)}{h} \frac{1 + \tan(x+h)\tan(x)}{1}[/itex]

    Is this what you mean?
     
    Last edited: Oct 28, 2004
  11. Oct 28, 2004 #10
    [tex]f'(x)=\lim_{h\,\rightarrow\,0}1+\tan(x+h)\tan(x)[/tex]

    What is [tex]\lim_{h\,\rightarrow\,0}\tan(x+h)[/tex]?
     
  12. Oct 28, 2004 #11

    NateTG

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    Nitpicking: Well, you also have to insure that [tex]\tan(x)[/tex] is defined. If [tex]x=\frac{\pi}{2}[/tex] the limit does not exist or is asymptotic.
     
  13. Oct 28, 2004 #12
    Nitpicking: you mean ensure :).
     
  14. Oct 29, 2004 #13
    Hm, I'm not seeing how to get to 1/cos^2x, but it doesn't matter, I'll just use f'(x)/g'(x).

    [itex]f(x) = \tan(x) \rightarrow (\frac{\sin(x)}{\cos(x)})' = \frac{(\sin(x))'\cos(x) - \sin(x)(\cos(x))'}{(\cos(x))^{2}} = \frac{\cos^{2}(x) + \sin^{2}(x)}{\cos^{2}(x)}[/itex]

    Here is another question though:
    Find the derivative of arctan(x), using arctan(tan(x)) = x.
    This also has me stumped.

    [itex]g'(x) = \arctan(x)[/itex]

    I know that:
    [itex]f'(x) = f'[g(x)] * g'(x)[/itex]

    But, I don't see what I can do with arctan(x), except for the obvious:

    [itex]\arctan(\arctan(\tan(x)))[/itex]
    which doesn't help at all.

    How would I proceed? I'm suppoed to use the derivative of a "functionsfunction", not sure what it's called in english.
     
    Last edited: Oct 29, 2004
  15. Oct 29, 2004 #14
    Say y = arctan(x) then tan(y) = tan(atan(x)). But tan(atan(x))=x for all x. So we have tan(y) = x deriving with respect to x on left side and right side we get:
    y' * (tan(y)^2 + 1) = 1 => y' = 1/(tan(y)^2+1). But what is tan(y)? Well tan(y)=x so the derivate of arctan(x) is 1/(x^2+1).
     
  16. Oct 29, 2004 #15
    Mmm I know to do it but with complex function, since

    [tex] \arctan{z} = {1 \over 2} i ( \ln(1 -iz) - \ln(1 + iz))[/tex]

    You easily obtain that

    [tex] {d \over dz} \arctan{z} = {1 \over 1 + z^2}[/tex]

    I don't know how to do in other way, without knowing the expression of arctan in more elemmental functions...
     
  17. Oct 29, 2004 #16
    Hm.
    [itex]y = \arctan(x)[/itex]
    [itex]\tan(y) = x[/itex]
    [itex]\arctan(\tan(x)) = x[/itex]
    This I understand.
    But, how do you get from:
    [itex]y = \arctan(x)[/itex]
    to
    [itex]tan(y) = \tan(\arctan(x))[/itex]

    Why isn't it, or wouldn't it be:
    [itex]tan(y) = \arctan(\tan(x))[/itex]
    ?
     
  18. Oct 29, 2004 #17

    NateTG

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    Take the tangent of both sides of the equation.
     
  19. Oct 29, 2004 #18
    Ah, of course :)
    so
    [itex]\tan(y) = \tan(\arctan(x)) = x[/itex]
    And:
    [itex]f'[g(x)] * g'(x)[/itex]

    [itex]f(x) = \tan(y)[/itex] and [itex] g(x) = y[/itex]

    [itex](tan(y))' * y' = x'[/itex]

    [itex]\frac{1}{\cos^2(x)} * 1 = 1[/itex]

    I'm not understanding this :(
    How do you get [itex]\frac{1}{\cos^2(x) + 1}[/itex]
    ?
     
    Last edited: Oct 29, 2004
  20. Oct 29, 2004 #19
    > Take the tangent of both sides of the equation.

    I don't know if that is too correct since the tangent is defined between [0, pi/2) so what if x = n*pi/2?
     
  21. Oct 29, 2004 #20

    NateTG

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    [tex]\tan (\arctan (x))=x[/tex]
    [tex]\frac{d}{dx} \tan (\arctan (x)) = \frac{d}{dx} x[/tex]
    [tex]\arctan'(x) \times \sec^2(\arctan(x))=1[/tex]
    [tex]\arctan'(x) = \cos^2(\arctan(x))[/tex]
    Now
    [tex]\cos(\arctan(x))=\frac{\pm 1}{\sqrt{1-x^2}}[/tex]
    (Consider, for example, a right triangle where the adjacent side is 1, the opposite side is [tex]x[/tex] and the hypotenuse is [tex]\sqrt{1+x^2}[/tex])
    So we can substitute that in:
    [tex]\arctan'(x)=(\frac{\pm 1}{\sqrt{1-x^2}})^2=\frac{1}{1+x^2}[/tex]
    the [tex]\pm[/tex] drops out because of the square.
     
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