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Derivatives of Log and Exponential Functions

  1. Dec 28, 2004 #1
    Hi i was wondering if someone could check my work for these couple questions:

    Find dy/dx (do not simplify)

    a)y=e^sin3x
    dy/dx= e^sin3x (cos3x)(3)
    =3cos3xe^sin3x

    b)y=5^(square rootx) x^2
    dy/dx=x5^(square rootx) (xln5+2)

    c)y=ln(x^2 / (2x+5)^3 )
    y=lnx^2 - ln(2x+5)^3
    dy/dx= 2x/x^2 - 6(2x+5)^2 / (2x=5)^3
    =2/x - 6/(2x+5)

    d)y=4log base 2 (square rootx+1)
    y=4logbase2 (x+1)^-1
    dy/dx= 4/lnbase2(square rootx+1)

    e)y=ln[x^2 - e^x / x^2 +e^x]
    y= ln(x^2 -e^x) - ln(x^2 +e^x)
    dy/dx= [2x -e^x / x^2 - e^x] - [2x +e^x / x^2+e^x]

    f)e^x^2 multiplied by y^3=x (isolate dy/dx)
    I am unsure of how to do this one, i have never seen one like this before, could someone show what i would do.

    g)Use logarthimic differentiation to find dy/dx if
    y=[e^x cosx / (square root x)]^5

    lny=5xlne +5lncosx -5/2lnx
    dy/dx= y[5x-5tanx-5/2x]
    dy/dx= =[e^x cosx / (square root x)]^5 [5x-5tanx-5/2x]

    h)A radioactive substance decays in such a way that the amount in grams present at time t years is given by A(t)=100e^-0.2t
    i)What is the initial amount of radioactive material, A(0)?
    A(0)= 100
    ii)Find the rate of decay function, A'(t).
    A'(t)=-20e^-0.2t
    iii)How mucgh radio active material is present when t-50 years? How fast is the material decaying at this time?
    A(50)=100e^-10
    =0.0045399
    A'(50)=-20e^-10
    =-0.0009079
    iv)At what time t is one half of the original substance remaining? What is the decay rate at this time?
    I am not sure what to do here also if someone could show me how id really appreciate it.
    *For this question and parts of the question i am not sure if some of my equations are correct as the numbers i am getting do not seem like it should be what they are.

    j)solve the logarithmic equation logbase3(x-3) + logbase3(x) = logbase3(4)
    I am also not sure what to do for this question.

    Thanks in advanced for the help! If someone could show me how to do those few questions and check my work on those ones cuz my text book does not show what the answers are or how to do some of them.
     
  2. jcsd
  3. Dec 28, 2004 #2

    dextercioby

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    This is gonna be a hell of a long post. :tongue2: This part is correct.

    [tex] \ln y=\sqrt{x}\ln5 +2\ln x;\frac{d\ln y}{dx}=\frac{1}{y}\frac{dy}{dx} [/tex]
    So:[tex]\frac{dy}{dx}=y\frac{d\ln y}{dx}=(5^{\sqrt{x}}x^{2})(\frac{\ln 5}{2\sqrt{x}}+\frac{2}{x})=...[/tex]

    Okay...

    I don't understand where is the square root and what's that argument for the [itex]\log_{2} [/itex]

    I don't follow you.If u have:
    [tex] y=\ln(x^{2}-\frac{e^{x}}{x^{2}}+e^{x}) [/tex],then the decomposition u found is wrong.It's correct iff u have
    [tex] y=\ln(\frac{x^{2}-e^{x}}{x^{2}+e^{x}}) [/tex]
    Then your result would be correct.However,the fact that you're not writing it in 'tex' misleads me.

    Here u got me completely lost.Is it
    [tex] e^{x^{2}}y^{3}=x [/tex]????????If so,then take natural log.out of both sides and get:
    [tex] x^{2}+3\ln y=\ln x [/tex].Differentiate wrt to 'x' to find
    [tex] 2x+\frac{1}{y}\frac{dy}{dx}=\frac{1}{x} [/tex].From there extract the derivative and substitute 'y' by the expression foud from the first equation.

    Okay.

    So far so good.
    You found that,initially,A(0)=100.The question is:find 't' from the eq.
    [tex] A(t)=100e^{-0.2t}=50 [/tex]


    Raise both sides of the eq. to the power 3.Use the property
    [tex] 3^{x+y}=3^{x}3^{y} [/tex]
    to find the solution.

    Daniel.
     
  4. Dec 28, 2004 #3
    Ya some of it is confusing as i dont know how to use that tex thing. For question d) the root is over x+1. And for question e) the equation is the second one you posted which as you said my answer is correct. f) is the equation you said so then i did what you said and this is the answer i get.

    dy/dx= (cube root of x-e^x^2)(1/x -2x)
    Is that correct?

    And for h)iv)
    50=100e^-0.2t
    ln0.5=-0.2t
    t=ln0.5/-0.2
    so then:
    A'(ln0.5/-0.2)=-20e^ln0.5
    Would i leave my answer at that, and is that correct?

    For question j) i still don't understand what you mean, could you please show me what the first line of the solution would be please. Thanks so much for the help!
     
  5. Dec 28, 2004 #4
    For (j), he means this:

    [tex]\log_b{xy} = \log_b{x}+\log_b{y}[/tex]
     
  6. Dec 28, 2004 #5

    dextercioby

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    d)[tex] y=4\log_{2}\sqrt{x+1} \Rightarrow 2^{y}=2^{4\log_{2}\sqrt{x+1}}=(2^{\log_{2}\sqrt{x+1}})^{4}=(\sqrt{x+1})^{4}=(x+1)^{2}[/tex]
    Take '\ln' from both sides of the eq.to find:
    [tex] y=\frac{2}{\ln 2}\ln(x+1)\Rightarrow \frac{dy}{dx}=\frac{2}{(\ln 2)(x+1)} [/tex]


    h)iv) [tex] t=5\ln 2 \Rightarrow A'(5\ln 2)=20e^{-0.2\cdot 5\cdot\ln 2}=\frac{20}{2}=10 [/tex]

    It's not correct what u've written.Mass cannot be negative.

    j)[tex]x(x-3)=4 \Rightarrow x=4 [/tex]
    The solution x=-1 implies logarithm in the base "3" from a negative number...

    Daniel.
     
    Last edited: Dec 28, 2004
  7. Dec 28, 2004 #6

    dextercioby

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    For f):
    [tex] \frac{dy}{dx}=y(\frac{1}{x}-2x)=(\sqrt[3]{x e^{-x^{2}}})(\frac{1}{x}-2x) [/tex]


    Daniel.
     
  8. Dec 28, 2004 #7
    Alright thanks i understand now! Thanks so much.
     
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