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Math Help - Implicit Differation and Chain Rule

  1. #1
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    Implicit Differation and Chain Rule

    First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
    The problem I need help with is:

    x^2 + xy - y^2 = 4 Solver for y'
    I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).

    Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

    Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

    Thank you!!!
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    Hello,

    Quote Originally Posted by PensFan10 View Post
    First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
    The problem I need help with is:

    x^2 + xy - y^2 = 4 Solver for y'
    I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).
    Your only mistake (that I found by analysing your answer, so sorry if you did it right..) is that the derivative of xy has to be taken with the product rule :

    (xy)'=(x)'y+x(y)'=y+xy'

    As soon as the variable intervenes in x, y, z, t, or whatever, you have to consider it as a function, not like a variable, so you're likely to use the product, quotient or chain rule =)

    Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

    Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

    Thank you!!!
    \sqrt{\frac{2x}{x^2+1.8}}=\sqrt{u(x)}

    The chain rule says that the derivative is \frac{u'(x)}{2 \sqrt{u(x)}}

    So calculate u'(x) with the quotient rule.

    Always do things step by step. It's pretty tricky for someone who is not used to differentiating to do it without taking u(x).

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    Thanks so much

    That was stupid of me to mess up the xy (product rule). Doh! Thanks for catching that mistake.

    Thanks a lot for the help!!!
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    One more quick question:

    So I solved chain rule problem out and got an answer of:

    (1)/(XSqrt(2x/x^2+1.8))

    Just curious if I did it correct. Once again, Thanks!
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    Quote Originally Posted by PensFan10 View Post
    So I solved chain rule problem out and got an answer of:

    (1)/(XSqrt(2x/x^2+1.8))

    Just curious if I did it correct. Once again, Thanks!
    Hmm; that's a weird one.

    What do you find as the derivative u'(x) ?
    Because I doubt it is \frac 2x /:

    Can you show your working so that I can tell you where the mistake(s) is(are) ?
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    Of course

    Okay, I see how I messed up.
    Heres my current solution attempt :P

    For u'(x) I have this:

    2x/(x^2+1.8)

    Quotient rule says: gf'-fg'/g^2

    When I plug my info in I got:

    (x^2+1.8)(2)-(2x)(2x)/(x^2+1.8)^2
    Simplified:
    2x^2+3.6-4x^2/(x^2+1.8)^2

    =-2x^2+3.6/(x^2+1.8)^2

    Now I plug that into u'(x)/(2Sqrt u(x))

    I got

    -2x^2+3.6
    -----------------
    (x^2+1.8)^2
    ------------------
    2Sqrt(2x/x^2+1.8)

    Might be just me, but I think I did something wrong again.
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    Quote Originally Posted by PensFan10 View Post
    Okay, I see how I messed up.
    Heres my current solution attempt :P

    For u'(x) I have this:

    2x/(x^2+1.8)

    Quotient rule says: gf'-fg'/g^2

    When I plug my info in I got:

    (x^2+1.8)(2)-(2x)(2x)/(x^2+1.8)^2
    Simplified:
    2x^2+3.6-4x^2/(x^2+1.8)^2

    =-2x^2+3.6/(x^2+1.8)^2

    Now I plug that into u'(x)/(2Sqrt u(x))

    I got

    -2x^2+3.6
    -----------------
    (x^2+1.8)^2
    ------------------
    2Sqrt(2x/x^2+1.8)

    Might be just me, but I think I did something wrong again.
    Ignoring the missing parentheses, this is correct

    Note that you can write -2x^2+3.6=2(-x^2+1.8) and simplify the fraction by 2

    The solution is :

    \frac{2(-x^2+1.8)}{2(x^2+1.8)^2 \sqrt{\frac{2x}{x^2+1.8}}}=\frac{-x^2+1.8}{(x^2+1.8)^2 \cdot \sqrt{2x} \cdot (x^2+1.8)^{\color{red}-\frac 12}}

    -----------------------------
    do you understand the red part ?

    \frac{1}{a^b}=a^{-b}, and \sqrt{a}=a^{\frac 12}
    -----------------------------

    Simplifying the powers by this rule : a^b a^c=a^{b+c}, we get :

    f'(x)=\frac{-x^2+1.8}{\sqrt{2x} \cdot (x^2+1.8)^{\frac 32}}

    Now it depends on how you want to display the result
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  8. #8
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    Awesome!

    Yes the red part was you moving the denominator to the numerator. Thanks so much and sorry about the missing parentheses. If I knew you, I would buy you a drink
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    Quote Originally Posted by PensFan10 View Post
    Yes the red part was you moving the denominator to the numerator. Thanks so much and sorry about the missing parentheses. If I knew you, I would buy you a drink
    Great, I was being thirsty !
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  10. #10
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    Quote Originally Posted by PensFan10 View Post
    First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
    The problem I need help with is:

    x^2 + xy - y^2 = 4 Solver for y'
    I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).

    Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

    Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

    Thank you!!!
    Alternatively for the second one, Let

    y=\sqrt{\frac{2x}{x^2+1.8}}\Rightarrow{y^2=\frac{2  x}{x^2+1.8}}

    So differentiating we get

    2y\cdot{y'}=\frac{2(x^2+1.8)-(2x)(2x)}{(x^2+1.8)^2}=\frac{-2(x^2-1.8)}{(x^2+1.8)^2}\Rightarrow{y'=\frac{-(x^2-1.8)}{y(x^2+1.8)^2}}

    and now remembering that y is our original equation we see

    y'=\frac{-(x^2-1.8)}{\sqrt{\frac{2x}{x^2+1.8}}(x^2+1.8)^2}

    EDIT:

    Same concept, just a little less messy to write
    Last edited by Mathstud28; June 7th 2008 at 01:07 PM.
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    okay

    i kind of see what you did. I understand squaring both sides and then taking the derivative. I got lost after you simplified.

    how did you go from 2(x^2+1.8)-(2x)(2x) to -2(x^2-1.8) (i left out the denominator to be less confusing)

    Second, is that actually supposed to be a 4.8 in the numerator in your final answer or is it a typo. *EDIT* never mind. i see you changed it. now that part makes more sense :P
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PensFan10 View Post
    i kind of see what you did. I understand squaring both sides and then taking the derivative. I got lost after you simplified.

    how did you go from 2(x^2+1.8)-(2x)(2x) to -2(x^2-1.8) (i left out the denominator to be less confusing)

    Second, is that actually supposed to be a 4.8 in the numerator in your final answer or is it a typo.
    That is a typo that I fixed, and secondly

    (2(x^2+1.8)-(2x)(2x)=2x^2+2(1.8)-4x^2=-2x^2-(-2)(1.8)=-2(x^2-1.8)
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  13. #13
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    haha

    Ha. Not sure how I didn't see that Thanks. I really like that way.
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