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Math Help - trig derivatives

  1. #1
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    trig derivatives

    Find the derivative and simplify in each case:

    f(x) = arctan (x - \sqrt{1+x^2})

    f(x) = e^{tan^{-1}(x^2)}

    f(t) = \frac{sin^{-1}(t)}{sin^{-1}(2t)}

    Thanks to all that helps : )
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  2. #2
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    I assume that {{\sin }^{-1}}x=\arcsin x.

    This is purely chain rule application. What have you done?
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  3. #3
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    i have the last 2 done and i cannot get the first one because i cannot get the derivative of the value within the brackets

    and yes arcsin x = sin^{-1} x

    1. ????

    2. f'(x) = \frac{2xe^{tan^{-1}x^2}}{1 + x^4}

    3. f'(x) = \frac{\sqrt{1 - 4t^2}}{2\sqrt{1 - t^2}}
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  4. #4
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    Okay, put h(x)=x-\sqrt{1+x^2} so that h'(x)=1-\frac{x}{\sqrt{1+{{x}^{2}}}}.

    Hence, we have f(x)=\arctan \big(h(x)\big)\implies f'(x)=\frac{h'(x)}{1+{{\big(h(x)\big)}^{2}}}. Now just remove the clutter.

    -----

    I didn't check if your other derivatives were okay but try this site to see it:

    solving derivatives step-by-step
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  5. #5
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    haha thats what i had done! : )
    thank you very much!
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  6. #6
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    so is the answer to my first one:

    \frac{\sqrt{1+x^2} - x}{2\sqrt{1+x^2}(x^2 - x\sqrt{1+x^2} + 1)}

    or am i completely wrong? lol
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  7. #7
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    Hello, qzno!

    The first one is challenging
    . . with some surprising simplification at the end.


    f(x) \:=\: \arctan \left[x - \sqrt{1+x^2}\right]
    Of course, we use the Chain Rule . . .

    f'(x) \;=\;\frac{1}{1 + \left[x-\sqrt{1+x^2}\right]^2} \cdot\bigg[1 - \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\bigg]

    . . . . = \;\frac{1}{1 + x^2 - 2x\sqrt{1+x^2} + 1 + x^2}\cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}}

    . . . . = \;\frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}}

    . . . . = \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}} . . . . which is what you got!


    Multiply by \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}

    . . \frac{1}{2((1+x^2 - x\sqrt{1+x^2})} \cdot\underbrace{\frac{\sqrt{1+x^2} -x}{\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}}

    . . = \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})}\cdot\overbrace{\frac{1+x^2 - x\sqrt{1+x^2}}{1+x^2}} . . . now reduce!

    . . = \;\frac{1}{2(1+x^2)}

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  8. #8
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    thank you!
    that helped me out so much : )
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