anonymous
  • anonymous
find the derivative of ln ((square root of x^2 +1)/(x(2x^3-1)^2))
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
\[\large \ln\left(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}\right)\] So this is what our problem looks like? Oh boy this one is gonna be a doozy. We'll have to apply the chain rule, and then the pruduct rule, and then the chain rule a few more times.
zepdrix
  • zepdrix
The derivative of natural log gives us,\[\large \left(\ln \heartsuit\right)'=\frac{1}{\heartsuit}\heartsuit'\]I wanted to use some besides x, `Whatever` the contents of the log may be, you stuff all of that into the denominator. The prime on the outside is due to the chain rule. we have to multiply by the derivative of the inside of the log. The little prime is to let us know we still need to differentiate it. So if we apply this idea to our problem, here's what we get! :O
zepdrix
  • zepdrix
\[\large \frac{d}{dx}\ln\left(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}\right)\quad = \quad\] \[\large \frac{1}{\left(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}\right)}\frac{d}{dx}\left(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}\right)\]

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anonymous
  • anonymous
looks like a lot of work... try simplifying \(\large ln(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}) \) using the properties of logs first...
zepdrix
  • zepdrix
Ah yes, that would be better -_- woops!

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