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Math Help - Differentiate functions

  1. #1
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    Differentiate functions

    1. Find dy/dx for each of the following. (You do not need to simplify your answers)
    a) y = loge 3x + 8x
    b) y = e^x cosec x
    c) y = e^4x+3/4x + 3
    d) y = √2x - 1

    2. If g(x) = 7x - 2x + 4x - 8 find the third derivative g'''(x)
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  2. #2
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    Quote Originally Posted by Kiwigirl
    1. Find dy/dx for each of the following. (You do not need to simplify your answers)
    a) y = loge 3x + 8x
    b) y = e^x cosec x
    c) y = e^4x+3/4x + 3
    d) y = √2x - 1

    2. If g(x) = 7x - 2x + 4x - 8 find the third derivative g'''(x)
    Hello,

    with most of your problems you have to use the chain rule. With b) I used the quotient rule and with c) you have to controle if I wrote the function correctly:

    to a):
    y=\ln(3x)+8x^3 \Longrightarrow y'=\frac{1}{3x} \cdot 3+3 \cdot 8x^2

    to b):
    y=\frac{e^x}{\cos(x)} \Longrightarrow y'=\frac{e^x*\cos(x)-e^x*(-\sin(x))}{(\cos(x))^2}

    to c):
    y=e^{4x}+\frac{3}{4x}+3 \Longrightarrow y'=e^{4x} \cdot 4+\frac{3}{4} \cdot \left(-\frac{1}{x^2} \right)

    to d):
    y=\sqrt[3]{2x}-1 \Longrightarrow y'=\frac{1}{3} \cdot (2x)^{-\frac{2}{3}} \cdot 2

    To 2.: You have a function of grade 3. You want to get the third derivative. That must be a constant. So g'''(x)=3!*7=42.

    Greetings

    EB
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  3. #3
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    Thanks earboth for your quick response I really appreciate it!
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