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Help with new calculus concept

  1. Sep 28, 2006 #1
    Hi all (it has been a while again),

    I have just started Universtiy and after my second lecture I was given a question sheet to complete in two weeks. I have completed 14 out of the 16 questions set, 2 of which I just haven't done yet and another will require a little more thought before I need some help.

    The question I am having a problem with is this one:

    4. Find f'(x) for each of the following f.
    (d) [tex] f(x) = x^{sinh2x} [/tex]

    I would give some sort of attempt, like you all expect, before I get any help but I am afraid I have never come across this sort of equation before. I really do not know where to start. I did think about letting u = sinh2x and doing a sustitution but I realised that would not work so I am really stuck. I can but assume I need some sort of rule or theorem that I have either not learnt or cannot find on the internet (and trust me, I have looked).

    Any help would be greatly appreciated.

    Cheers for now,

    The Bob (2004 ©)
     
  2. jcsd
  3. Sep 28, 2006 #2

    matt grime

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    It's the chain rule. You've met the chain rule?
     
  4. Sep 28, 2006 #3
    [tex]\log_{e}f = ?[/tex]
     
  5. Sep 28, 2006 #4
    Is it truely as simple as the chain rule?

    Also I have never done a f'(x) question by using log base e before I start.

    So even if it is as simple as the chain rule (which has caused myself to be quiet embarrassed) what does the right hand side become? I am not going to pretend I understand at this stage, despite feeling I should be able to.

    Cheers,

    The Bob (2004 ©)
     
  6. Sep 28, 2006 #5
    Use the rule log(a^b) = b log(a), then differentiate the equation, using the the product and chain rules, with respect to x.
     
  7. Sep 28, 2006 #6
    Let me have a quick go.

    Does sinhx need to be substituted for some lettter (e.g. u)? Then I can differentiate with respect to u (for f(x)) and with respect to x (for u).

    So is

    [tex]f'(x) = 2 cosh2x . sinh2x . x^{(sinh2x - 1)}[/tex]

    ?

    Cheers,

    The Bob (2004 ©)
     
  8. Sep 28, 2006 #7
    But then I have a [tex]log_e f'(x)[/tex], do I not?

    The Bob (2004 ©)
     
  9. Sep 28, 2006 #8
    Ok, so will I have

    [tex] \frac{d(log_e y)}{dx} = \frac{sinh2x}{x} + (2cosh2x)(log_e x)[/tex]

    ?

    The Bob (2004 ©)
     
  10. Sep 28, 2006 #9
    [tex]\frac{d(\log f(x))}{dx} = \frac{1}{f(x)}f'(x)[/tex]

    This is yet another application of the chain rule...
     
  11. Sep 28, 2006 #10
    You can't use the same rule for differentiating x^n, with n constant. sinh(2x) is a function of x.
     
  12. Sep 28, 2006 #11
    I did think as much but felt it was, stupidly, worth a try. :frown:

    The Bob (2004 ©)
     
  13. Sep 28, 2006 #12

    matt grime

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    Forget logs. There is no need to use them.

    The derivative of exp(x) is exp(x). The derivative of sinh(2x) is 2cosh(2x). Just apply the chain rule

    (f(g(x)))' = g'(x)f'(g(x)).
     
  14. Sep 28, 2006 #13
    But in the OP's case, the base is not e.
     
  15. Sep 28, 2006 #14

    matt grime

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    Doh. stupid me.
     
  16. Sep 28, 2006 #15
    You can either take the log of both sides (as suggested by others), or just rewrite it as

    [tex]f(x) = e^{\sinh (2x)\ln{x}}[/tex]

    and apply the chain rule directly.
     
  17. Sep 28, 2006 #16
    I was wondering why you said to just use the chain rule.

    I can see how, from this, I can get a solution:

    [tex]f(x) = e^{\sinh (2x)\ln{x}}[/tex] Let u = sinh(2x).ln(x)

    [tex]\Rightarrow f(x) = y = e^u[/tex]

    [tex]\Rightarrow f'(x) = \frac{dy}{dx} = e^u[/tex]


    Chain rule says that [tex]\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}[/tex]

    [tex]\frac{du}{dx}[/tex] requires the product rule so [tex]\frac{du}{dx} = ln(x) \cdot 2 \cosh (2x) \ + \ \frac{\sinh (2x)}{x}[/tex]

    Thus [tex]\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = (e^{\sinh (2x)\ln{x}}) \times (ln(x) \cdot 2 \cosh (2x) \ + \ \frac{\sinh (2x)}{x})[/tex]

    [tex]\Rightarrow \frac{dy}{dx} = x^{\sinh (2x)} \times (ln(x) \cdot 2cosh(2x) \ + \ \frac{sinh(2x)}{x})[/tex]

    Is this now correct?

    Cheers,

    The Bob (2004 ©)

    P.S. It would appear that the LaTex is not working too well so here is the final equation in text:

    dy/dx = (x^(sinh(2x))) x ((ln(x)).(2cosh(2x)) + (sinh(2x)/x))
     
    Last edited: Sep 28, 2006
  18. Sep 28, 2006 #17
    Apart from the line where you said, f'(x) = dy/dx = e^u, you've are correct. It should've been e^u.du/dx, but since you got chain rule part right, I'll let you off. ;)
     
  19. Sep 29, 2006 #18
    Aha.... ok. :biggrin: Thanks for everyone's help. I must admit that was a technique (rather than a principle or theorem) that I had not done before, as simple as it was.

    I do hope in future the interent, or the large text book I have, will yield more answers before I embarasse myself here.

    Thanks again everyone,

    The Bob (2004 ©)
     
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