What are the first and second derivatives of $f (x) = ln (x^{2 x + 1})$ $f(x)=ln(x^(2x+1) )$ ?

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What are the first and second derivatives of $f (x) = ln (x^{2 x + 1})$ $f(x)=ln(x^(2x+1) )$ ?

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Answer 1 · 2015-11-27T22:42:45

First derivative: $\frac{d y}{d x} = 2 ln (x) - \frac{1}{x} + 2$ $(dy)/(dx) =2ln(x)-1/x+2$

2nd derivative: $\frac{2}{x} + \frac{1}{x^{2}}$ $2/x + 1/(x^2)$ Did not have time to do the $\frac{d^{2}}{d x}$ $(d^2)/dx$ properly so just gave the answer!

Explanation:

Given: $y = ln (x^{2 x + 1})$ $y=ln(x^(2x+1))$

Write as : $y = (2 x - 1) ln (x)$ $y=(2x-1)ln(x)$

Using standard form $\frac{d y}{d x} = v \frac{d u}{d x} + u \frac{d v}{d x}$ $(dy)/(dx)=v (du)/dx+u(dv)/(dx)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let $u = 2 x - 1 \to \frac{d u}{d x} = 2$ $u=2x-1 -> (du)/dx=2$
Let $v=ln(x) ->color(white)(...) (dv)/dx = 1/x$

Then $(dy)/(dx) = ln(x)(2)+(2x-1)(1/x)$

$(dy)/(dx) = 2ln(x)+(2x)/x-1/x$

$(dy)/(dx) =2ln(x)-1/x+2$

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What are the first and second derivatives of $f (x) = ln (x^{2 x + 1})$ $f(x)=ln(x^(2x+1) )$ ?

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Explanation:

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What are the first and second derivatives of f(x)=ln(x^(2x+1) ) f(x)=ln(x2x+1)f(x)=ln(x^(2x+1) ) ?

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What are the first and second derivatives of $f (x) = ln (x^{2 x + 1})$ $f(x)=ln(x^(2x+1) )$ ?