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Simplify the function and find the derivative

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Simplify the function and find the derivative;
    (Attached)

    2. Relevant equations

    I've had a go at the question (see below) trying to use trigonometric identities.

    3. The attempt at a solution

    I somehow arrived at 1/7 - 3/2sec^2, but I'm not sure that this is correct!
     

    Attached Files:

    Last edited: Jun 3, 2013
  2. jcsd
  3. Jun 3, 2013 #2

    HallsofIvy

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    Show what you did. I assume you know that [itex]e^{ln(x)}= x[/itex]. Do you know that [itex]cosh^2(x)- sinh^2(x)= 1[/itex]?
     
  4. Jun 3, 2013 #3
    Thanks for your reply. I've had another attempt and attached the working. I've tried to write neatly... :D
     

    Attached Files:

  5. Jun 3, 2013 #4

    CAF123

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    You are making it so much more difficult for yourself. Before differentiating, simplify the expression as much as possible. Also is that part at the end ##\sqrt{tan^3}## or ##\sqrt{tan3}##? I don't think the steps you wrote were correct.
     
  6. Jun 3, 2013 #5
    Sorry it should be the second one... no power... (sqrt(tan3)). I'll have a go at simplifying.
     
  7. Jun 3, 2013 #6
    Remember, the term in the square brackets is a function of x, and it's multiplied by another function of x. That mean you will have to use the product rule, which you didn't appear to do. Similarly, the multiplying function is a quotient of two functions of x, so you'll have to use the quotient rule there too. It look like you just replaced each individual function by it's derivative, which isn't correct.
     
  8. Jun 3, 2013 #7
    Thank you for your help! I've just read your post DimReg so I'm guessing my third attempt isn't correct either! *sigh*
     

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  9. Jun 3, 2013 #8

    CAF123

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    You are differentiating the components of the expression individually which is not correct. You have to use product rule. Note also if you fully simplify the expression in the square brackets, this comes out very simple and you need only apply product rule once.
     
  10. Jun 3, 2013 #9
    Ahhhh, does that mean that within the brackets it will simplify to 27+1? Because of the rule that you previously posted?
     

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  11. Jun 3, 2013 #10

    CAF123

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    Yes, so now simplify the quotient a bit further and then just apply product rule once. (or quotient rule). I say you can use either because you can apply quotient on the form ##f/g ## or product on the form ##fg^{-1}##
     
  12. Jun 3, 2013 #11
    I'm really grateful for your help, I'm terrible at Maths and this question is doing my head in!

    So from the attached, if I figure out how to calculate the end result of the quotient rule (in red)... then I still apply the product rule for 4*(result from quotient rule)? I wasn't sure if the quotient rule applied if one of the functions wasn't a function of x.
     

    Attached Files:

  13. Jun 3, 2013 #12
    Thanks, I'll keep trying to evaluate the quotient rule... was I right to assume previously that the derivative of sqrt(tan3) =0?
     
  14. Jun 3, 2013 #13
    I'm stuck again... with the quotient rule, is any of this right?
     

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  15. Jun 3, 2013 #14

    CAF123

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    Sorry, but I don't think so. You seem to be making the expression so much more complicated than what it should be with the right amount of simplification. Also, there is some algebra errors there too: $$\frac{a - b}{a} \neq -b\,\,\,\,\,\text{e.g}\,\,\,\,\,\frac{2-1}{2} \neq -1$$

    You have the following: $$\frac{28 \cosh^2x}{7 \sinh 2x} - \sqrt{\tan 3}$$ As you said above, the derivative of the sqrt term is zero (since it is a constant) so simplify the quotient further.
     
  16. Jun 3, 2013 #15
    :( thanks. Is the way I wrote the rule out correct? I'll go back from there. Any other hints?
     
  17. Jun 3, 2013 #16

    CAF123

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    I don't think so - there appears to be constants missing and you have three terms on the numerator.


    Did you see the edit to my last post? Maybe try to start from there and simplify the quotient further using more hyperbolic identities.
     
  18. Jun 3, 2013 #17
    Okay.... pleaaaaaaase tell me I'm closer now?! I hadn't seen your edits before my last attempt so hopefully I might be on the right track?

    **sorry it didn't rotate, I think I'm getting sleepy, 2.22am here!**
     

    Attached Files:

  19. Jun 3, 2013 #18

    CAF123

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    It's nearly perfect! - you can just make one more simplication: $$\frac{1}{\sinh^2x} = \csch ^2x$$ and you're done.

    Goodnight.
     
    Last edited: Jun 3, 2013
  20. Jun 3, 2013 #19
    YAY! Thank you so much!! You've been so helpful.

    Goodnight :)
    Tam
     
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