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Confused on these two derivative problems

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  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data

    a. k(t) = (sqrt(t+1))/(2t+1)


    b. y = (3^(x^2+1))(ln(2))


    3. The attempt at a solution
    For the first problem, I know I use the quotient rule for derivatives (L)(DH)-(H)(DL)/((L)^2)
    which would go to: ((2t+1)(1/(2sqrt(t+1)) - (sqrt(t+1))(2))/((2t+1)^2) I get stuck here, maybe it's the algebra but I don't know how to factor all this down to get a solution.

    For the second problem I was able to get the answer online but I don't understand the answer.
    It goes:

    y = (3^(x^2+1))(ln(2))
    refine y = e^((ln(3))(x^2+1)) ln(2)
    y' = [2xln(3)ln(2)]3^(x^2+1)

    I don't understand where the e or ln(3) comes from. Or why it's put as an exponent. Also, why does the ln(2) stay stationary? Then e disappears as x^2+1 is broken into its derivative, the ln(3) comes down while 3^(x^2+1) is still put on the side. Could someone please explain this process works.

    Thank you
     
  2. jcsd
  3. Aug 6, 2015 #2
    Take the LCM - it won't be simplified too much - but that's the way it's going to stay .
    This is just the manipulation of the original question in order to convert into terms which you can solve by chain rule . Do you know basic logarithmic properties ?
    You're only raising 3x2+1 to e .
    Do you know the chain rule ?

    Hope this helps .
     
  4. Aug 6, 2015 #3

    SteamKing

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    A suggestion:

    Rather than make up complicated expressions for derivatives, learn what notation is used normally and follow that:

    y = f(x)

    dy/dx = y' = f'(x) (use the ' to indicate differentiation w.r.t. x. The expression f'(x) is read, "f-prime of x".)

    then, expressing the quotient rule in prime notation,

    y = f(x) / g(x)

    y' = (f(x)g'(x) - f'(x)g(x)) / [g(x)]2
     
  5. Aug 6, 2015 #4

    SteamKing

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    When working problems involving taking derivatives, I find it easier to convert expressions like "sqrt (x)" or "√x " to exponential form, so that "sqrt(x) = √x = x1/2 ".
     
  6. Aug 6, 2015 #5

    HallsofIvy

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    You should know that [itex]a^x= e^{ln(a^x)}= e^{x ln(a)}[/itex]. As result of that, the derivative of [itex]a^x[/itex] is [itex] a^x ln(a)[/itex] and the derivative of [itex]a^{f(x)}[/itex] is [itex]f'(x)a^{f(x)} ln(a)[/itex].
     
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