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Help with my homework please

  1. Mar 22, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to just differentiate these 2
    1)[​IMG]

    2)[​IMG]

    And use implicit differentiation to fidn dy/dx on these 2
    3)[​IMG]

    4)[​IMG]


    I took a crack at both of them the 1st two i couldn't really complete, and the last two I'm not really confident on them. Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 22, 2008 #2
    Since you didn't do either of the first 2 correctly:

    For 1) You didn't apply the chain rule right
    For 2) You didn't apply the multiplication rule at all
    For 3) What's [itex]\frac{d}{dx}(6)[/itex]?
    For 4) You did fine until you didn't divide through totally properly on the second to last step.

    You need to slow down when doing these problems.
     
    Last edited: Mar 22, 2008
  4. Mar 22, 2008 #3
    [tex]\frac{d}{dx}(\sin x)=x'\cos x[/tex]
     
  5. Mar 22, 2008 #4
    would the third one be

    y'=-y^2-2xy/(x^2*2y)


    for the last one would it be

    y'= 1-2y/2xy?
     
  6. Mar 22, 2008 #5
    You factored a common term of y' incorrectly. Review your Algebra, it's not too late!!!
     
    Last edited: Mar 22, 2008
  7. Mar 22, 2008 #6
    thanks..that helps a lot..


    ok reviewed it

    3) y'=-y^2-2xy/x^2*2yx

    4) y'=1-2y/2x2y
     
    Last edited: Mar 22, 2008
  8. Mar 22, 2008 #7
    Let's do a review ... factor and solve for y:

    [tex]y+3x+zy=xy[/tex]
     
  9. Mar 22, 2008 #8
    sorry math and physics is not my forte

    i guess the review problem would be
    y=3x/xz?
     
    Last edited: Mar 22, 2008
  10. Mar 22, 2008 #9
    [tex]a(x+y+z)=ax+ay+az[/tex]

    Your mistake is that, you're not factoring out the common term ...

    [tex]y-xy+zy=-3x[/tex]

    Common term y

    [tex]y(1-x+z)=-3x[/tex]

    [tex]y=-\frac{3x}{1-x+z}[/tex]

    Math wasn't my fav. subject either, but it's definitely useful so deal with it.
     
  11. Mar 22, 2008 #10
    Notice on your 2nd to last line for your last problem ... you divided only the 1st 2 terms ... it has to be EVERYTHING!
     
  12. Mar 22, 2008 #11
    damn..can't believe it was that simple i forgot about it maybe the "factor" was a dead giveaway.

    So if i applied that to problems 3 & 4

    3) y'=-y^2-2xy/x^2+2yx

    4) y'=1-2y/2x+2y-1

    please tell me i did it right..
     
  13. Mar 22, 2008 #12
    Yes, those are good now.
     
  14. Mar 22, 2008 #13
    Did you rework the first one?

    When I do the chain rule of trigonometric functions, like in problem one, I always think of the regular function, sin x, with the stuff stuffed in, and then the derivative of the stuff.

    As rocomath said d/dx[sin u] = (cos u) u'

    G(x) = sin 3x and G'(x) = 3 cos 3x

    u = 3x cos u = cos 3x
    u' = d/dx [3x] = 3

    Also, with trigonometry, it helps to pay attention to the parentheses.

    Studying these may help quite a bit:

    y = cos 3x^2 = cos (3x^2)

    y' = (-sin 3x^2) (6x) = -6x sin 3x^2

    y = (cos 3)x^2

    y' = (cos 3) (2x) = 2x cos 3

    y = cos (3x)^2 = cos(9x^2)

    y' = (-sin 9x^2)(18x) = -18x sin 9x^2

    y = cos^2 x = (cos x)^2

    y' = 2(cos x) (-sin x) = -2 cos x sin x
     
  15. Mar 22, 2008 #14
    yeah, i know how to do chain rule..

    derive outside, leave inside, then derive outside, but since theres it's 3pi(x)/2 would i use the quotient rule or what. I know 3 pi/2 is a constant.
     
  16. Mar 22, 2008 #15
    What is the derivative of kx if k is a constant? It's just k, no quotient rule needed for that. The chain rule is the derivative of the outside function times the derivative of the inside function.
     
  17. Mar 22, 2008 #16

    Well, the chain rule is where the mistake is being made. You don't have to worry about the quotient rule.

    Take a look at:

    cos (3x^2)

    y' = (-sin 3x^2) (6x) = -6x sin 3x^2


    IN problem one you posted, you have as the first "term" sin ((3 * pi * x) / 2) as the minuend.

    So it's cos ((3 * pi * x)/2)

    times the derivative of 3 * pi * x / 2.

    You pull out the constant, which leaves you with the constant times the derivative of x, which is one.

    So the first term is ((cos (3 * pi * x)/2) * 3 * pi)/2

    You can do the other one the same way.

    It helps to study example of the Chain Rule to master it. The Chain Rule is so important that you will seldom again differentiate any function without using it.
     
  18. Mar 22, 2008 #17
    yeah i wrote it down wrong..
     
  19. Mar 22, 2008 #18
    yeah i've had this it's whats next is what i have the problem on..

    cos(3*pi*x/2)(3*pi/2)-sin(3*pi*x/2)(3*pi/2)

    do i just leave it like that? or should i rearrange the constant to the front.
     
    Last edited: Mar 22, 2008
  20. Mar 22, 2008 #19
    Doesn't matter ... it's still the same. However, it's customary to write the constant in the front. Also, in this case, they both have a common term of that same constant, so factoring it out would be the best choice.

    [tex]y'=x'(\cos x-\sin x)[/tex]
     
  21. Mar 22, 2008 #20
    Ok got it...

    On the second one I used the chain rule and ended up with:

    -2(2x-5)^-2 * (12x-30)(x^2-5x)^5

    and now i'm stuck yet again.
     
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