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Need help with finding derivative

  1. Sep 30, 2007 #1
    The first problem is...
    y = 1/2 e^x - 3 sinx
    Derivative = 1/2e^x - 3 cosx * 1
    Is that right? I wasn't sure about the 1/2.

    The second problem I have is...
    f(x) = (2e^x)/(x^2 + 1)
    I am not sure what the derivative of 2e^x is....is it just 2e^x?
    (x^2 + 1)(2e^x) - (2e^x)(2x)/(x^2 + 1)^2

    We don't have to simplify.
     
  2. jcsd
  3. Sep 30, 2007 #2
    the first one's right.

    and so is the second one.

    if you have 2e^x the 2 is just a constant so it's the same as 2*d(e^x)/dx
     
  4. Sep 30, 2007 #3
    :O even on the test? either way it's good practice to simplify.
     
  5. Sep 30, 2007 #4
    Thank you guys. No he doesn't want us to simplify on tests either because it is easier to grade for him...hehe.
     
  6. Sep 30, 2007 #5
    So for 2 sinxcosx the derivative would be...
    (2cosxcosx)*(-sinx)
     
  7. Sep 30, 2007 #6
    And for x^2 tanx^2 you would treat that as 3 seperate derivatives?
    (2x) * (sec^2x^2) * (2x)
     
  8. Sep 30, 2007 #7
    what trig identity could you use?

    2sinxcosx=? by doing that, your derivative becomes much easier.
     
  9. Sep 30, 2007 #8
    3? i only see the product rule

    [tex]x^{2}\tan^2{x}[/tex]

    correct?
     
  10. Sep 30, 2007 #9
    Trig identity is sin2x right?
    (cos2x) * (2)
     
  11. Sep 30, 2007 #10
    Yea thats right. I thought you took the derivative of x^2 then tan x^2 and x^2
     
  12. Sep 30, 2007 #11
    i think you mean you will need to do the chain rule

    if so, that is correct
     
  13. Sep 30, 2007 #12
    Oh yes the chain rule, sorry. So this is correct...
    (2x) * (sec^2x^2) * (2x)
     
  14. Sep 30, 2007 #13
    no! what book are you using?

    [tex]x^{2}\tan^{2}x[/tex]

    can be re-written as

    [tex]x^{2}(\tan{x})^2[/tex]

    how do you differentiate and apply the chain rule to tanx?
     
  15. Sep 30, 2007 #14
    Oh okay forgot about that...
    2x(tanx)^2 * (sec^2x)^2
    do you also multiply by 2(tanx) or no.
     
  16. Sep 30, 2007 #15
    differentiating and applying the chain rule to (tanx)^2

    [tex]2\tan{x}\sec^2{x}[/tex]
     
  17. Sep 30, 2007 #16
    Okay so the final answer would be...
    (2x(tanx)^2) * 2(tanx sec^2x)
     
  18. Sep 30, 2007 #17
    it shouldn't be times, it should be + ? (product rule)
     
  19. Sep 30, 2007 #18
    I thought you said to use the chain rule?
     
  20. Sep 30, 2007 #19
    were you giving me the answer to x^2tan^2x or just tan^2x?

    if you were giving the overall answer, then it's still incorrect.
     
  21. Sep 30, 2007 #20
    the tan x^2
    overall answer being...
    (2x) + 2(tanx sec^2x)
     
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