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Homework Help: Derivatives with pictures (problems 3-4)

  1. Nov 20, 2008 #1
    2gv1mz6.jpg

    6oevrq.jpg

    Help me out please ! for #4 I'm preeeetty sure i did it correctly but it wouldn't hurt having someone double check it. for #5 , I'm not sure if i took the second derivative correctly

    Will be on for the next couple hours trying to understand calculus ... :(

    Thanks in advance ^^
     
  2. jcsd
  3. Nov 20, 2008 #2

    Dick

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    You are pretty good on 4), the other two are an absolute disaster. I don't think you quite get the chain rule. Let's just take a part. What's d/dx(x*y)? Use the product rule also.
     
  4. Nov 20, 2008 #3
    Thanks for answering

    d/dx(xy) = (y)+(x)(dy/dx)

    but i dont see where im supposed to apply the chain rule
    i'm having trouble with the second part...

    d/dx(xy^3)= (y^3) + (3y^2)*(dy/dx)*x

    d/dx(10)=0

    then I got [y^3+x*3y^2*dy/dx] + [y+x*dy/dx] = 0

    EDIT: wait is d/dx (xy^3) = y^3+x3y *dy/dx
     
  5. Nov 20, 2008 #4

    Dick

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    Now you've got it right. Good job. That's different from what you wrote in the post. Now what is dy/dx?
     
  6. Nov 20, 2008 #5
    [(y^3+x3y*dy/dx)] = [-(y+x*dy/dx)]

    how do i cancel the dy/dx ?


    since i'm almost done, i'll try to work on the other problem too. btw, is my #4 correct?

    #5 f(x)= -2ln[sin(x)]

    f'(x)= -2cos(x)/sin(x) --> -2cos(x)/sin(x) . b/c ln(u(x))' = u'(x)/u(x)

    f"(x)= quotient rule ? sin(x)*2sin(x)+cos(x)(2cos(x)) / sin(x)^2

    Thanks a lot ! ^^
     
  7. Nov 20, 2008 #6

    Dick

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    You don't cancel dy/dx. You solve for it. Move all the dy/dx parts to one side of the equation, factor dy/dx out and move everything else to the other side. And, yes, I think you are doing ok with part 4 as well. You could also say f'(x)=-2*cot(x) and go from there.
     
  8. Nov 20, 2008 #7
    Haha I think I'm doing this wrong but ...

    #3. -y^3-y / x3y / x and then i have to use tangent line equation after this. but i'll do it after you confirm its correct or wrong


    #5. I took your advice f'(x)-2*cot(x)

    f"(x) = -2csc^(x) is the answer ?
     
  9. Nov 20, 2008 #8

    Dick

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    I get +2csc(x)^2, but you might want to check that, it's late here. For the first one, the numerator looks good, but what's going on the denominator? x3y?
     
    Last edited: Nov 21, 2008
  10. Nov 21, 2008 #9
    Yup, you were correct but it's f"(x) 2csc^2(x)

    and for this one...


    Wait, was the one in the "EDIT" correct or incorrect?

    but if its correct, -y^3 -y / 3y after cancelling out the x's
     
  11. Nov 21, 2008 #10

    Dick

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    You can't cancel that x, now can you?
     
  12. Nov 21, 2008 #11
    i dont understand this at all . its confusing me , i'm not even sure if i'm supposed to look at

    1. d/dx (xy^3) = y^3+x3y *dy/dx

    or

    2.d/dx(xy^3)= (y^3) + (3y^2)*(dy/dx)*x

    assuming its 1. this time i got (-y^3-y)(x) / x3y .............


    after i use the implicit differentiation, to find the tangent line to the curve , i use point slope formula right ?
     
  13. Nov 21, 2008 #12

    Dick

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    It's 2. d/dx(y^3)=3y^2*dy/dx. Chain rule. You had it right when you said "I got [y^3+x*3y^2*dy/dx] + [y+x*dy/dx] = 0". Solve that for dy/dx. I'll repeat what I already said. "Move all the dy/dx parts to one side of the equation, factor dy/dx out and move everything else to the other side." By dividing by it. Like this: if x+y=x*dy/dx-y*dy/dx, dy/dx=(x+y)/(x-y).
     
    Last edited: Nov 21, 2008
  14. Nov 21, 2008 #13
    okay this time i got -y^3-y / 3y^2 . damn i hope this is right .

    with all these "y's" how can i use the point slope formula to get y=mx+b for question #3 (the one dealing with implicit differentiation)
     
  15. Nov 21, 2008 #14

    Dick

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    Look at your correct differentiation in the preceding post. Where did the term x*dy/dx go? And what happened to the x in x*3*y^2? You'll finally use (x,y)=(5,1) and figure out dy/dx which is m in your point slope form. Can you show exactly how you solved for dy/dx?
     
  16. Nov 21, 2008 #15
    okay i am getting really confused .

    here's what i did so far :

    an09yu.jpg

    I thought you told me to just leave the 2 dy/dx's on the left side and then I could cancel out the x if i bring it to the right side x/x = 1
     
  17. Nov 21, 2008 #16

    Dick

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    Take your second line from the bottom. That becomes (x*3y^2+x)*dy/dx=(-y^3-y). So dy/dx=(-y^3-y)/(x*3y^2+x). You can't move the 3y^2 over all by itself, and you can't just 'cancel' the x. There's nothing to 'cancel' it. Those aren't calculus problems, those are algebra problems.
     
  18. Nov 21, 2008 #17
    Thanks. the next thing I have to do is find the tangent line to this...

    i need to find the slope first. Do i need to take derivative again? I don't think so because i already did implicit differentiation...

    so the next step is to plug in the x value (5,1) so i plug in 5 ... into all the x's.

    I haev a question, do I get y alone on one side of the equation and then plug in the x ?
     
  19. Nov 21, 2008 #18

    Dick

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    Yes, find the slope first. That's dy/dx. You have a formula for dy/dx. Put x=5 and y=1 into it.
     
  20. Nov 21, 2008 #19
    after plugging in, i get dy/dx = -1/10 which is the slope

    then i use point slope formula y-y0 = m (x -x0)

    y-(1) = -1/10x-1/2

    i get y= -1/10x + 1/2 so m= -1/10 and b= 1/2 ?
     
  21. Nov 21, 2008 #20

    Dick

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    Your problems with algebra are killing you. -1/10*(x-5)=(-1/10)*x+1/2 NOT (-1/10)*x-1/2. You should also practice checking yourself. If you think y=(-1/10)*x+1/2, check it. If you put in x=5, you should get y=1. You don't.
     
    Last edited: Nov 21, 2008
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