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Homework Help: Finding the derivative and expressing in factored form

  1. Dec 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y = (x-2)exp3 x sqrt 2x-1

    2. Relevant equations

    3. The attempt at a solution

    I got a final answer of (x-2)^2(2x-1)^-1/2(13x-14)

    What do you guys think? Is this correct?
  2. jcsd
  3. Dec 23, 2008 #2


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    I can't read the original question. Is it supposed to be: [tex]y=(x-2)e^{3\sqrt{2x-1}}[/tex]?
  4. Dec 23, 2008 #3
    Yeah thats supposed to be the question. I dont know how to type it up any better than that sorry.
  5. Dec 23, 2008 #4


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    Well I realised I can't read your answer as well. But I don't think it's correct since the exponential term doesn't appear anywhere.
  6. Dec 23, 2008 #5
    Sorry, I used the ^ for the exponent

    It should read (x-2)exp2(2x-1)exp-1/2(13x-14)
  7. Dec 23, 2008 #6
    Again im sorry, i made a mistake. The question should read

    y = ((x-2)exp3) (sqrt 2x-1)
  8. Dec 23, 2008 #7
    Take log on both sides and then try differentiating it
  9. Dec 23, 2008 #8


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    Still ambiguous.
    [tex]y= (x-2)^3 \sqrt{2x-1}[/tex]
    or [tex]y= (x-2)^3(\sqrt{2x}-1)[/tex]?

    If the first, write it as [itex](x-2)^3 (2x-1)^{1/2}[/itex] and use the chain rule and product rule.

    If the second, write it as [itex](x-2)^3(\sqrt{2}x^{1/2}- 1)[/itex] and again use the chain rule and product rule.
  10. Dec 23, 2008 #9
    What is the definition of the chain rule with the product rule? They give an example of that in the manual but its not very clear.
  11. Dec 23, 2008 #10


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    The product rule says
    d(uv)/dx = (du/dx)v+ u(dv/dx)

    The chain rule is d(u(v))/dx= du(v)/dv (dv/dx)
  12. Dec 23, 2008 #11
    How come I have to use both of those rules? Wouldnt 1 suffice?
  13. Dec 23, 2008 #12
    Applying the product rule first you realize you'll be using the chain rule automatically when calculating the derivatives: Take the derivative of the first term,expression,etc and multiply it by the second term. Thats the first part of your product rule. Then ADD: the derivative of the second term multiplied by the first term. (Or whichever order you prefer)

    (f*g)' = f'*g + g'*f

    The ' , just means the derivative. f' means the derivative of f, and g' means the derivative of g.

    Now for the chain rule. When you take the derivative of the first term you notice you need to take the derivative of (x-2)^3, bring the 3 down infront, minus 1 from the exponent, and multiply by the derivative of the inside of (x-2), which is 1:

    3(x-2)^2 * 1

    A similar operation will be done for the next derivative, if you have problems seeing the "power rule/chain rule" with the square root, just change the squareroot to the power of 1/2. The chain rule is used when you have functions nested inside other functions. So it need not be a power. Say you have cos(5x), the 5x is nested inside the cos(). So taking the derivative you would get -sin(5x)*5. Generally it's just a good idea to remember the power rule and the chain rule applied to trigonomic functions. If you need more examples or maybe another way of describing it, just google power rule or chain rule examples or search the forums. If you really wanna know where it comes from, try picking up an introductory analysis textbook or just search for a proof.

    Quick example, try doing this one first. Its a little easier: The derivative of:
    [tex](5x-1)^2 (4+2x)^3
    =(5x-1)^2*3(4+2x)^2 2 [/tex] [tex]+[/tex] [tex] (4+2x)^3 *2(5x-1) 5[/tex]
  14. Dec 23, 2008 #13
    Do I always use the product rule, followed by the chain rule when finding the deriviative of 2 factors?
  15. Dec 23, 2008 #14
    Nope, if you had to differentiate cos(x)*e^x you can just use the product rule (you are still using the chain rule when you are taking the derivative of the arguments of your function but since they're just x and you know the derivative of x is 1 and multiplying by 1 is ... well you get the idea)
  16. Dec 23, 2008 #15
    OKay, so lets work this through with the question I have from above.

    I got (2x-1)^-1/2(x-2)^2(7x-5) for a final answer?
    Last edited: Dec 23, 2008
  17. Dec 24, 2008 #16
    That's wrong unfortunately. Use the chain rule on (x-1)3 and (2x-1)1/2 seperately. The combine them and use the product rule.

    e.g by using chain rule: (d/dx)(x-1)3 = 3(x-1)2
  18. Dec 24, 2008 #17
    The first factor should be (x-2)^3

    Okay, I used the chain rule on the 2 factors separately. I got 3(x-2)^2 and (2x-1)^-1/2

    Now I dont knoow how to use the product rule with factors that have exponents in them.

    Using the product rule I get an equation of 3(x-2)^2 -1/2(2x-1)^-3/2 + (2x-1)^-1/2 6(x-2)
  19. Dec 28, 2008 #18
    Im still having trouble with this question here.

    Using the product rule and the chain rule together I have

    3(x-2)^2(2x-1)^1/2 + (2x-1)^-3/2 (x-2)^3

    so from here I need to factor out some of this,

    (x-2)^2(2x-1)^-3/2 [x-2/3(2x-1)^1/3

    where do i go from here?
    Last edited: Dec 28, 2008
  20. Dec 29, 2008 #19
    I'll work this out for you trying to detail each step, usually I'm in favor of suggesting hints rather than giving solutions but it has been long enough where you might benefit more from reading each step in detail.

    All right assuming this is the problem [tex](x-2)^3 \sqrt{2x-1}[/tex].

    Basically we have a product of 2 functions

    [tex] f(x) = (x-2)^3[/tex] and [tex] g(x) = \sqrt{2x-1} [/tex].

    We know the product rule for derivatives tells us:

    [tex] (f(x) g(x))^{'} = f^{'}(x)g(x) + f(x)g^{'}(x) [/tex]

    (where ' denotes derivative)

    So in our case we get the following:

    [tex] f(x) = (x-2)^3 \Rightarrow f'(x) = 3(x-2)^2[/tex] and [tex] g(x) = \sqrt{2x-1} \Rightarrow g'(x) = \frac{1}{\sqrt{2x-1}} [/tex]

    Putting this together into our formula we get:

    [tex] 3(x-2)^{2}\sqrt{2x-1} + \frac{(x-2)^{3}}{\sqrt{2x-1}} = \frac{(x-2)^{2}}{\sqrt{2x-1}}\left(3(2x-1) + (x-2)\right) = \frac{(x-2)^{2}}{\sqrt{2x-1}}(7x-5) [/tex]

    Personally I have found that most Calculus students suffer not because they do not understand the Calculus portion of the material but because their Algebra is rather spotty. It might help in your studies if you went back and reviewed some of the old material.
  21. Dec 30, 2008 #20
    Thats kind of frustrating cause I got that as the final answer in one of my previous posts, with the exception of not bringing the 2x-1^-1/2 down, and the one guy said that was wrong and I should try it another way.
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