Results 1 to 5 of 5

Thread: implicit differentiation

  1. #1
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    679
    Thanks
    1

    implicit differentiation

    find $\displaystyle dy/dx$ by implicit differentiation

    $\displaystyle \tan^3{(xy^2+y)} = 1$

    didn't know how to deal with the $\displaystyle \tan^3$


    the answer is:
    $\displaystyle \frac
    {1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)}
    {3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    For the differentiation of tan^3, use the chain rule !

    Here are the first steps :

    We know that the derivative of $\displaystyle \tan^3(x)=3\cdot\tan^2(x)\cdot \frac{1}{\cos^2(x)}$, by the chain rule.

    So the derivative of $\displaystyle \tan^3(x)(z)$, where $\displaystyle z=xy^2+y$ will be $\displaystyle 3\cdot\tan^2(z)\cdot \frac{1}{\cos^2(z)}\cdot\frac{dz}{dx}$, by the chain rule.

    Now I bet you can differentiate $\displaystyle z=xy^2+y$ with respect to x (not forgetting that y is a function of x)
    Last edited by Moo; Mar 16th 2010 at 03:22 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    666
    Thanks
    2
    Awards
    1
    Quote Originally Posted by bigwave View Post
    find $\displaystyle dy/dx$ by implicit differentiation

    $\displaystyle \tan^3{(xy^2+y)} = 1$

    didn't know how to deal with the $\displaystyle \tan^3$


    the answer is:
    $\displaystyle \frac
    {1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)}
    {3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}$
    Hi,

    Just recall the chain rule... first you take the power and forget everything else:

    $\displaystyle thing^3(stuff) = 1$

    Now we differentiate:

    $\displaystyle 3thing^2(stuff)thing' = 1$

    Now, we must differentiate the stuff inside, so that we get:

    $\displaystyle 3*thing^2(stuff)*(thing)'*(stuff)' = 1$

    So, if we let:

    $\displaystyle thing = tan()$
    $\displaystyle stuff = xy^2 + y$
    $\displaystyle thing' = sec^2()$
    $\displaystyle stuff' = y^2 + 2xyy' + y'$

    Plug this all in to get:

    $\displaystyle 3tan^2(xy^2 + y)[sec^2(xy^2 + y)(y^2 + 2xyy' + y')] = 1$

    Now, we can divide everything without a y' to the other side:

    $\displaystyle y^2 + 2xyy' + y' = \frac{1}{3tan^2(xy^2 + y)sec^2(xy^2 + y)}$

    Now, we move the y squared over and factor out the y':

    $\displaystyle y'(2xy + 1) = \frac{1}{3tan^2(xy^2 + y)sec^2(xy^2 + y)} - y^2$

    Now we divide through by 2xy + 1:

    $\displaystyle y' = \frac{1}{3(2xy + 1)tan^2(xy^2 + y)sec^2(xy^2 + y)} - \frac{y^2}{2xy + 1}$

    And if you get common denominators it becomes the answer you have... Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    679
    Thanks
    1
    also, can we consider $\displaystyle tan^3(x)$ the same as $\displaystyle (tan(x))^3$ to apply $\displaystyle d/dx$
    Last edited by bigwave; Mar 16th 2010 at 03:39 PM. Reason: edit I saw the correction
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by bigwave View Post
    how did you get this?

    I thot $\displaystyle \tan^3{(x)} = 3\tan^2{(x)}\sec^2{(x)}$

    also, can we consider $\displaystyle tan^3(x)$ the same as $\displaystyle (tan(x))^3$ to apply $\displaystyle d/dx$
    Sorry, I edited while you were apparently posting. Got messed up with some derivatives

    As for your second sentence, yes, $\displaystyle \tan^3(x)=(\tan(x))^3$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Jul 26th 2010, 05:24 PM
  2. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 10th 2010, 05:58 PM
  3. implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 20th 2009, 10:34 AM
  4. implicit differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 12th 2009, 03:15 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 21st 2008, 08:12 PM

Search Tags


/mathhelpforum @mathhelpforum