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Homework Help: Finding the second derivative

  1. Apr 12, 2013 #1
    1. Find the second derivative of the following function



    2. [itex] f(x) = x\sqrt{5-x}, f'(x) = \frac{10-3x}{2\sqrt{5-x}}[/itex]



    3. [itex] f"(x)=\frac{-3*2\sqrt{5-x}-(10-3x)(2\sqrt{5-x})*-x}{(2\sqrt{5-x})^{2}} [/itex] Here is where I get lost.
     
  2. jcsd
  3. Apr 12, 2013 #2
    [itex] f"(x)=\frac{-3*2\sqrt{5-x}-(10-3x)(2\sqrt{5-x})*-x}{(2\sqrt{5-x})^{2}} [/itex] Here is where I get lost. [/b][/QUOTE]

    There is a slight error in this step. When you differentiated the denominator in the second term, you have not subtracted 1 from the 1/2 power. This is what I got:

    [tex] \frac{-3(2\sqrt{5-x})-(10-3x)(2)\frac{1}{2}(5-x)^{-1/2}(-1)}{(2\sqrt{5-x})^2} [/tex]
     
  4. Apr 13, 2013 #3

    Mark44

    Staff: Mentor

    Also, it's probably simpler to write f'(x) as a product rather than a quotient, and write the radical in exponent form. That way you can use the product rule to get f''(x). I almost always prefer to use the product rule over the quotient rule, because the latter is a bit more complicated, making it easier to make mistakes.

    f'(x) = (1/2)(10 - 3x)(5 - x)-1/2
     
  5. Apr 13, 2013 #4
    I know a lot of sites have this as the first derivative [itex] \sqrt{5-x}(1-\frac{x}{2(5-x)})[/itex] I am wondering what happens to the square root on the bottom? I am just wondering what is easier to work with for the second derivative.
     
  6. Apr 13, 2013 #5
    Using the product rule, I have [itex] \frac{1}{2} (10-3x)*\frac{-1}{2}(5-x)^{-3/2}*-1-3(5-x) = \frac{1}{4}\frac{10-3x-15+3x}{(5-x)^{3/2}} [/itex] Let me know if I did anything wrong.
     
  7. Apr 13, 2013 #6
    I see what I did wrong, I forgot to multiply 1/2 all the way through.
     
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