1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivatives with pictures (problems 3-4)

  1. Nov 20, 2008 #1
    [​IMG]

    [​IMG]

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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 :

    [​IMG]

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?