Results 1 to 7 of 7

Math Help - dy/dx of tan^2(x^3)

  1. #1
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Cool dy/dx of tan^2(x^3)

    find \frac{dy}{dx} of \tan^2(x^3)

    I know that \frac{dy}{dx}\tan{x} = \sec^2{x} and probably to use the chain rule
    but didn't know how to deal with the powers?

    the ans is: 6x^2\tan(x^3)\sec^2{(x^3)}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by bigwave View Post
    find \frac{dy}{dx} of \tan^2(x^3)

    I know that \frac{dy}{dx}\tan{x} = \sec^2{x} and probably to use the chain rule
    but didn't know how to deal with the powers?

    the ans is: 6x^2\tan(x^3)\sec^2{(x^3)}
    Note that \tan^2(x^3)=(\tan(x^3))^2.

    Thus, when differentiate this, you will need to use chain rule twice.

    Can you proceed?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    next step

    so if

    then would my next step be:

    2(\tan{x^3})(\frac{-3x^2}{\cos^2{x^3}})

    this looks like it is heading towards the answer.

    not sure what you mean by doing chain rule twice
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    y = (\tan x^3)^2

    Make u =  \tan x^3 \implies y =  u^2 \implies \frac{dy}{du}=2u

    Applying the chain rule again separately on u =  \tan x^3

    Make w = x^3 \implies u = \tan w \implies \frac{du}{dw}= \sec^2w \implies \frac{dw}{dx}= 3x^2

    \frac{du}{dx}= \sec^2w \times 3x^2

    \frac{dy}{dx} = 2u \times \sec^2w \times 3x^2 = 6x^2 \tan x^3 \sec^2 x^3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by bigwave View Post
    so if

    then would my next step be:

    2(\tan{x^3})(\frac{-3x^2}{\cos^2{x^3}}) where did the minus come from, bigwave?

    this looks like it is heading towards the answer.

    not sure what you mean by doing chain rule twice
    \frac{1}{cos^2\left(x^3\right)}=sec^2\left(x^3\rig  ht)

    x^3=u=f(x)

    w=tanu=f(u)

    y=w^2=f(w)

    \frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{du}\frac{du}{  dx}

    =2wsec^2u3x^2=2[tanu]sec^2\left(x^3\right)3x^2=6x^2tan\left(x^3\right)s  ec^2\left(x^3\right)

    Ah! pickslides! where'd you come from?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Archie Meade View Post
    Ah! pickslides! where'd you come from?
    Didn't mean to spoil the party...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    someday I want be famous like you guys are. look a all those rep points..

    a vain hope probably

    the neg sign just slipped in by itself.... I certainly didn't put it there.

    thnxs much... now I see the light... of u and w substitution
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum