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Math Help - Logarithmic Differentiation Problem and Natural Log Derivative/Integral

  1. #1
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    Logarithmic Differentiation Problem and Natural Log Derivative/Integral

    Stuck on logarithmic differention, after taking the natural log of each side:

    1. dy/dx for y=[(x)squareroot(2x+7)] / [x-2]^3

    2. Y=ln[(squareroot(x^2+3))/(x^4)]

    3. Evaluate the Integral(antiderivative) : [2x-1]/[x-1]dx

    Need Help on these, final tomorrow. Thanks.
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  2. #2
    Super Member Showcase_22's Avatar
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    y=\frac{x \sqrt{2x+7}}{(x-2)^3}

    This is just an application of the quotient rule:

    \frac{dy}{dx}=\frac{(x-2)^3 \left( \sqrt{2x+7}+\frac{x}{ \sqrt{2x+7}}\right)}{(x-2)^6}

    \frac{dy}{dx}=\frac{\sqrt{2x+7}+\frac{x}{\sqrt{2x+  7}}}{(x-2)^3}

    I'll leave it to you to tidy it up!

    The second one follows directly from the first. Remember that \frac{d}{dx} ln(f(x))=\frac{f'(x)}{f(x)}. Did I write this in your other thread?

    The final one is much easier to just solve:

    \int \frac{2x-1}{x-1} \ dx= \int \frac{x+x-1}{x-1} \ dx = \int 1+\frac{x}{x-1} dx

    From here it can be split up:

    \int 1 \ dx + \int \frac{x}{x-1} \ dx

    The first integral is easy, the second one is harder.

    Let u=x-1 \Rightarrow \ \frac{du}{dx}=1

    x+ \int \frac{u+1}{u} \ du+K

    x+\int 1+\frac{1}{u} \ du+K

    x+ u+ ln(u)+C

    x+x-1+ln(x-1)+C \Rightarrow \ 2x-1+ln(x-1)+C

    There's probably an easier of doing it that that, this method was just the first one I thought of.

    Good luck in your finals!
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