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Thread: inverse trigonometric differentiation

  1. #1
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    inverse trigonometric differentiation

    find the derivative of f(x) = arcsin(-x)
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  2. #2
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    Quote Originally Posted by myoplex11 View Post
    find the derivative of f(x) = arcsin(-x)
    The standard derivative is that the derivative of arcsin(x) is:

    1/(sqrt(1 - x^2)); it can be proved with some effort.

    The derivative of arcsin(-x) then is just -1/(sqrt(1 - x^2)).
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  3. #3
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    Draw a triangle such that $\displaystyle \sin(y)=x$. Add the hypotenuse and then differentiate both sides of this equation, remembering y is a function of x.
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  4. #4
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    Hello, myoplex11!

    No logs are involved in this problem . . .


    Find the derivative of: $\displaystyle f(x) \:= \:\arcsin(-x)$

    As AfterShock pointed out, there is a standard formula for arcsine.

    If you are expected to derive this formula, you can do it . . .


    We have: .$\displaystyle y \:=\:\arcsin(x)$

    Take the sine of both sides: .$\displaystyle \sin y \:=\:x$

    Differentiate implicitly: .$\displaystyle \cos y\cdot\frac{dy}{dx} \:=\:1 \quad\Rightarrow\quad \frac{dy}{dx}\:=\:\frac{1}{\cos y}$

    Since $\displaystyle \sin y \,= \,\frac{x}{1} \,= \,\frac{opp}{hyp}$, .Pythagorus tell us that: $\displaystyle adj \,= \,\sqrt{1 - x^2}$

    . . Hence: $\displaystyle \cos y \,= \,\frac{\sqrt{1 - x^2}}{1} \:=\:\sqrt{1 - x^2}$

    And the formula becomes: .$\displaystyle \frac{dy}{dx} \:=\:\frac{1}{\sqrt{1-x^2}} $

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  5. #5
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    The composite function,
    $\displaystyle \sin (\sin^{-1} (x))=x$
    Now there is a theorem that says that $\displaystyle \sin^{-1}(x)$ is differenciable function so use the chain rule,
    $\displaystyle [\sin^{-1}(x)]'\cos (\sin^{-1} (x))=1$
    Thus,
    $\displaystyle ;\sin^{-1}(x)]'=\frac{1}{\cos (\sin^{-1}) (x)}=\frac{1}{\sqrt{1-x^2}}$

    So to find,
    $\displaystyle \sin^{-1}(-x)$ derivative use what AfterShock said.
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