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Math Help - implicit differentiation

  1. #1
    Super Member bigwave's Avatar
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    implicit differentiation

    find dy/dx by implicit differentiation

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

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


    the answer is:
    \frac<br />
{1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)}<br />
{3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}
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  2. #2
    Moo
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    Hello,

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

    Here are the first steps :

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

    So the derivative of \tan^3(x)(z), where z=xy^2+y will be 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 z=xy^2+y with respect to x (not forgetting that y is a function of x)
    Last edited by Moo; March 16th 2010 at 03:22 PM.
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  3. #3
    Super Member Aryth's Avatar
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    Quote Originally Posted by bigwave View Post
    find dy/dx by implicit differentiation

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

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


    the answer is:
    \frac<br />
{1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)}<br />
{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:

    thing^3(stuff) = 1

    Now we differentiate:

    3thing^2(stuff)thing' = 1

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

    3*thing^2(stuff)*(thing)'*(stuff)' = 1

    So, if we let:

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

    Plug this all in to get:

    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:

    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':

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

    Now we divide through by 2xy + 1:

    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.
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  4. #4
    Super Member bigwave's Avatar
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    also, can we consider tan^3(x) the same as (tan(x))^3 to apply d/dx
    Last edited by bigwave; March 16th 2010 at 03:39 PM. Reason: edit I saw the correction
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  5. #5
    Moo
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    Quote Originally Posted by bigwave View Post
    how did you get this?

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

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

    As for your second sentence, yes, \tan^3(x)=(\tan(x))^3
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