Differentiating Inverse Trigonometric Functions

Key Questions

  • What is the derivative of y=arccos(x)?

    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}

    Wataru · · Sep 14 2014
  • What is the derivative of f(x)=tan^-1(x) ?

    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)

    Truong-Son N. · 20 · Jul 1 2015
  • What is the derivative of y=cot^{-1}(x)?

    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.

    Amory W. · 8 · Sep 14 2014
  • What is the derivative of f(x)=sin^-1(x) ?

    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}

    Wataru · 1 · Sep 11 2014

Questions

Differentiating Trigonometric Functions