Logarithmic Differentiation Problem and Natural Log Derivative/Integral

**PMoNEY23** · April 14th 2009, 11:34 PM

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.

**Showcase_22** · April 15th 2009, 02:51 AM

$y=\frac{x \sqrt{2x+7}}{(x-2)^3}$

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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