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Math Help - determining the derivative of exponental functions

  1. #1
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    determining the derivative of exponental functions

    how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.

    determining the derivative of exponental functions-untitled.bmp

    Thx in advance to anyone who helps.
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  2. #2
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    Hello, math619!

    You're scary . . .

    First of all, these are not exponential functions.

    Second, it's not 1n
    It's ln\\ which stands for "natural log".

    And you're expected to know the derivative of a log function.
    (Knowing Log Properties and the Chain Rule couldn't hurt.)

    It sounds like you missed a few classes.

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  3. #3
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    Quote Originally Posted by math619 View Post
    how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.

    Click image for larger version. 

Name:	untitled.bmp 
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ID:	1555

    Thx in advance to anyone who helps.
    As Soroban says its \ln not 1n. That is its the natural logarithm, or log
    to the base e. That means if:

    y=\ln(x),

    then:

    e^y=x

    Then the property that you need to find the derivatives of your functions is (you will also need the chain rule):

    \frac{d}{dx}\ln(x)=\frac{1}{x}

    RonL
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, math619!

    You're scary . . .

    First of all, these are not exponential functions.

    Second, it's not 1n
    It's ln\\ which stands for "natural log".

    And you're expected to know the derivative of a log function.
    (Knowing Log Properties and the Chain Rule couldn't hurt.)

    It sounds like you missed a few classes.

    Lol, you make me chuckle Soroban.

    Any way, lets see what we can do.


    Quote Originally Posted by math619 View Post
    how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.

    Click image for larger version. 

Name:	untitled.bmp 
Views:	27 
Size:	6.5 KB 
ID:	1555

    Thx in advance to anyone who helps.
    But yes, 1n is definitely a first for me.

    a.) y = ln(x^2 + 3x + 4)

    The derivative of ln(x) is 1/x.

    But, we have a composition of functions. Thus, we need to take the derivative and multiply it, too, of the inside.

    Thus, 1/(x^2 + 3x + 4) * 2x + 3 = (2x + 3)/(x^2 + 3x + 4)

    b.) You need to clarify this one:

    y = log(x^2) or y = log(x)^2?

    Also, log base 10 I am assuming?

    c.) y = ln((x^2)/(3x - 4)^3)

    1/((x^2)/(3x-4)^3) * derivative of inside

    = ((3x - 4)^3)/(x^2) * quotient rule

    Recall the quotient rule: (f(x)/g(x))' = (g(x)*f'(x) - f(x)*g'(x))/(g(x)^2)

    ((3x-4)^3*2x - [x^2*3(3x-4)^2*3])/[((3x-4)^3)^2]

    I'll let you simplify: (-3x^2 - 8x)/(3x - 4)^4

    Thus,

    ((3x - 4)^3)/(x^2)*(-3x^2 - 8x)/(3x - 4)^4

    Simplified: (-3x - 8)/(3x^2 - 4x)

    Or you can expand it: 2/x - 9/(3x - 4)

    Both are equivalent.
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