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Homework Help: Derivatives problem

  1. Oct 8, 2005 #1
    Hey Everyone,

    I cannot believe I am asking questions as stupid as this, but I know I am doing these wrong and want to correct my bad habits. My questions are on derivatives, something I used to be really good at, I'm hoping it will start to come back to me soon. Anyway here they are along with my work:

    1) Differentiate square root2 *u + squareroot(3u)

    For this one I used the product rule for the first section because you had root 2 multiplied by the constant 2, and for the second I just took the derivative as normal. I ended up getting an answer of:
    (squareroot2 + u)/(3/2 square rootu). Sorry if this seems unclear, I don't know how to get the square root sign to show up on the computer.

    2) Differentiate (u^-2 + u^-3)(u^5 - 2u^2)

    For this I used to product rule of course and these were my steps:

    =(u^-2+u^-3)(5u^4-4u) + (u^5-2u^2)(-2u^-3-3u^-4)
    =3u^2-2u+2u^-2

    I don't know where I went wrong on this one, or if I even did go wrong, but something just doesn't seem right to me.

    3) y=(square root x-1)/(square root x+1)


    For this one I'm not really sure how to go about showing my work on the computer so I will just write my answer. I used the quotient rule to get it.

    y'=(-0.5x-x^-0.5)/(x^0.5+1)^2

    Again, I do not know where I'm going wrong or if I did I'm just unsure of my answer.

    4) f(x) = (ax+b)/(cx+d)

    For this one I treated the a, c and d as constant values. Then I used the quotient rule to differentiate. My biggest problem with this one is that the answer seems to long to be right.
    I think it might be a bit easier to show my work on this one:

    =((cx+d)(x)-(ax+b)(x))/(cx+d)^2
    =((x)(cx-ax+d-b))/(cx+d)^2

    Thanks a lot for any help you can give guys, I know I can get this stuff eventually it just takes me awhile to get a good grasp of it sometimes. Sorry for my lack of work shown, I really don't know how to show it on the computer, and I was worried any work I did show would make it even more confusing. Thanks a lot!
     
  2. jcsd
  3. Oct 8, 2005 #2

    mezarashi

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    Homework Helper

    Quite alot of problems here, and I'm afraid I don't have alot of time with me right now. I'll highlight some things I see.

    Question 1:
    Square root 2 should be treated as a constant a. Then the derivative of ax would be a due to the power rule. The derivative of terms summed together is their derivatives separately added together.

    Question 2:
    Your product rule looks fine. I think somewhere there your algebra failed you.

    Question 3:
    You may be getting confused due to the square roots. Replace the square root by the power representation ^-1/2, so square root (x+1) would be (x+1)^1/2. Keep in mind as well that 1/x = x^-1. You should use the chain rule in full form if you get confused easily.

    u = f(x)
    du = f'(x)dx
    du/dx = f'(x)

    dy/dx = du/dx x dy/du

    Question 4:
    I think you made a blunder when you did d/dx(ax + b ) = x. Wrong! The derivative is a! Redo the quotient rule keeping this in mind.
     
  4. Oct 8, 2005 #3
    1) Ok, I really don't know what I was thinking for that first question. I changed my method and I got (2u^3/2+square root 3)/(2square rootu)

    2) For the second I made an error in expanding out the brackets, my revised answer is 3u^2 + 2u^-2 + 2u

    3) For this one I think you misunderstood the question due to my inability to write this out on the computer. I'll try writing it out again:

    ((square rootx)-1)/((square root x)+1)

    4) Once again, I don't know what I was thinking here. My new answer now makes more sense to me, it is (da-bc)/(cx+d)^2

    Well, I hope that fixes them, but I'm still unsure of my answers. If anyone could tell me whether I am on the right track or not I would really appreciate their efforts.
     
  5. Oct 8, 2005 #4
    For 1 you mean [tex](\sqrt{2})u + \sqrt{3u}[/tex] right?

    If so, then you must remember that the first part, square root of 2, is a constant, then for the second part you can do the chain rule.

    For 2, you got the same answer I got. For this problem I would have been lazy and expanded the whole thing, and then take the derivative of each part.
     
    Last edited: Oct 8, 2005
  6. Oct 8, 2005 #5
    Yes that is what I mean for question 1. I suppose the new answer I posted in the previous post is still wrong then? If so I don't really know how to go about fixing it.
     
  7. Oct 8, 2005 #6
    Hmm, I guess I can not understand what your answer is.
    I got: [tex]\sqrt{2} + \frac{\sqrt{3}}{2\sqrt{u}} [/tex]
     
    Last edited: Oct 9, 2005
  8. Oct 9, 2005 #7
    Ok, I will try it again and see if I can manage to get what you did, there is more than likely a mistake in mine somewhere. Do my others looks ok now? Thanks again.

    *edit* I'm wondering if part of the problem with our answers being different, is that I put everything under a common factor of 2squarerootu. I may still have made an error though, I'm going to have to check that out.
     
    Last edited: Oct 9, 2005
  9. Oct 9, 2005 #8
    *****bump*****
     
  10. Oct 9, 2005 #9

    mezarashi

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    Question 1's solution from mattmns is okay and your Question 4. I haven't verified the rest.
     
  11. Oct 9, 2005 #10
    OK, I made a bit of an edit in one of my above posts but I'm not sure if that fixes the issue or not. Thanks again!
     
  12. Oct 9, 2005 #11
    I just realized I made an extremely stupid mistake in the first question so that is now fixed. The others still need a bit of help though.
     
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