Differentiating Inverse Trigonometric Functions

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Derivatives of the Inverse Trig Functions

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Key Questions

  • Most people remember this
    #f'(x)=1/{sqrt{1-x^2}}#
    as one of derivative formulas; however, you can derive it by implicit differentiation.

    Let us derive the derivative.
    Let #y=sin^{-1}x#.

    By rewriting in terms of sine,
    #siny=x#

    By implicitly differentiating with respect to #x#,
    #cosy cdot {dy}/{dx}=1#

    By dividing by #cosy#,
    #{dy}/{dx}=1/cosy#

    By #cosy=sqrt{1-sin^2y}#,
    #{dy}/{dx}=1/sqrt{1-sin^2y}#

    By #siny=x#,
    #{dy}/{dx}=1/sqrt{1-x^2}#

  • By Implicit Differentiation,
    #y'=-1/sqrt{1-x^2}#.

    Let us look at some details.
    #y=cos^{-1}x#

    by rewriting in term of cosine,
    #Rightarrow cos y=x#

    by implicitly differentiating with respect to #x#,
    #Rightarrow -sin y cdot y'=1#

    by dividing by #-sin y#,
    #y'=-1/sin y#

    by the trionometric identity #sin y =sqrt{1-cos^2y}#,
    #y'=-1/sqrt{1-cos^2y}#

    by #cos y =x#,
    #y'=-1/sqrt{1-x^2}#

  • I seem to recall my professor forgetting how to deriving this. This is what I showed him:

    #y = arctanx#

    #tany = x#

    #sec^2y (dy)/(dx) = 1#

    #(dy)/(dx) = 1/(sec^2y)#

    Since #tany = x/1# and #sqrt(1^2 + x^2) = sqrt(1+x^2)#, #sec^2y = (sqrt(1+x^2)/1)^2 = 1+x^2#

    #=> color(blue)((dy)/(dx) = 1/(1+x^2))#

    I think he originally intended to do this:

    #(dy)/(dx) = 1/(sec^2y)#

    #sec^2y = 1+tan^2y#

    #tan^2y = x -> sec^2y = 1+x^2#

    #=> (dy)/(dx) = 1/(1+x^2)#

  • The answer is #y'=-1/(1+x^2)#

    We start by using implicit differentiation:

    #y=cot^(-1)x#
    #cot y=x#
    #-csc^2y (dy)/(dx)=1#
    #(dy)/(dx)=-1/(csc^2y)#
    #(dy)/(dx)=-1/(1+cot^2y)# using trig identity: #1+cot^2 theta=csc^2 theta#
    #(dy)/(dx)=-1/(1+x^2)# using line 2: #cot y = x#

    The trick for this derivative is to use an identity that allows you to substitute #x# back in for #y# because you don't want leave the derivative as an implicit function; substituting #x# back in will make the derivative an explicit function.

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