Results 1 to 3 of 3

Math Help - Inverse Function Derivative

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    11

    Inverse Function Derivative

    Please help I can't get these 8!

    a) d/dx cos^-1(x^2+x)
    b) d/dx (cos(x^2+x))^-1
    c) d/dx sin^-1(ln(x)-x^3)
    d) d/dx sin(ln(x)-x^3)^-1
    e) d/dx ln(cos^-1(x))
    f) d/dx ln((cos x)^-1)
    g) d/dx exp(sin^-1(x))
    h) d/dx exp((sin x)^-1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by keemariee View Post
    Please help I can't get these 8!






    a) d/dx cos^-1(x^2+x)
    b) d/dx (cos(x^2+x))^-1
    c) d/dx sin^-1(ln(x)-x^3)
    d) d/dx sin(ln(x)-x^3)^-1
    e) d/dx ln(cos^-1(x))
    f) d/dx ln((cos x)^-1)
    g) d/dx exp(sin^-1(x))

    h) d/dx exp((sin x)^-1)
    \frac{D[arcsin(u(x))]}{dx}=\frac{u'(x)}{\sqrt{1-u^2(x)}}

    I hope that helps

    I will do three

    D) \frac{D[arcsin(\ln(x)-x^3)]}{dx}....I think that is what it says


    So we would get by the above formula \frac{(ln(x)-x^3)'}{\sqrt{1-(\ln(x)-x^3)^2}}=\frac{\frac{1}{x}-3x^2}{\sqrt{1-\ln^2(x)+2\ln(x)x^3-x^6}}

    H) \frac{D[e^{arcsin(x)}]}{dx}=(arcsin(x))'e^{arcsin(x)}=\frac{e^{arcsin(x)  }}{\sqrt{1-x^2}}

    and

    \frac{D[\ln(arccos(x))]}{dx}=\frac{(arcos(x))'}{arcos(x)}=\frac{\frac{-1}{\sqrt{1-x^2}}}{arcos(x)}=\frac{-1}{\sqrt{1-x^2}arcos(x)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by keemariee View Post
    Please help I can't get these 8!







    a) d/dx cos^-1(x^2+x)
    b) d/dx (cos(x^2+x))^-1
    c) d/dx sin^-1(ln(x)-x^3)
    d) d/dx sin(ln(x)-x^3)^-1
    e) d/dx ln(cos^-1(x))
    f) d/dx ln((cos x)^-1)
    g) d/dx exp(sin^-1(x))

    h) d/dx exp((sin x)^-1)
    I'll give hints, but I won't solve them:

    a) Recall that \frac{d}{dx}cos^{-1}u=-\frac{d}{dx}sin^{-1}u=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}

    b) note that \left(cos(x^2+x)\right)^{-1}\neq cos^{-1}(x^2+x)

    c) Recall that \frac{d}{dx}sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}

    d) note that \left(sin(ln(x)-x^3)\right)^{-1}\neq sin^{-1}(ln(x)-x^3)

    e) Apply chain rule. It should be easy now, since I gave you the derivative of cos^{-1}u

    f) Apply chain rule. Note that \left(cos(x)\right)^{-1}\neq cos^{-1}(x)

    g) Apply chain rule. It should be easy, since I gave you the derivative of sin^{-1}u.

    h) Apply chain rule. Note that \left(sin(x)\right)^{-1}\neq sin^{-1}(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. derivative of inverse function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 17th 2010, 09:42 AM
  2. Derivative of inverse function
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 7th 2009, 08:33 PM
  3. inverse function derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 3rd 2007, 06:32 PM
  4. inverse function derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 5th 2007, 01:16 AM
  5. inverse function derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 4th 2007, 09:21 AM

Search Tags


/mathhelpforum @mathhelpforum