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i set up my problem as:

1/2(d/dx ln(x(x+2) - d/dx ln(2x+1)- d/dx ln(3x+2)

this setup doesn't seem right can someone confirm that this is correct or incorrect

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    $\begingroup$ You have sufficient reputation that you should have learned by now to use mathjax. $\endgroup$ Nov 1, 2019 at 17:55
  • $\begingroup$ You have mismatched parentheses. Also, any particular reason it doesn't seem right? $\endgroup$ Nov 1, 2019 at 18:29
  • $\begingroup$ if you're satisfied with you're answer, then please click the green check to close the post. Thanks! $\endgroup$ Nov 5, 2019 at 13:22

1 Answer 1

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This should be enough to get you on the right track...

$$y = \sqrt{ \frac{x(x+2)}{(2x+1)(3x+2)} }$$

$$\ln y = \ln \sqrt{ \frac{x(x+2)}{(2x+1)(3x+2)} }$$

$$\ln y = \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg)^\frac{1}{2}$$

$$\ln y = \frac{1}{2} \cdot \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg)$$

$$\frac{d}{dx} \big[ \ln y \big]= \frac{d}{dx} \Bigg[ \frac{1}{2} \cdot \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg) \Bigg]$$

$$\frac{d}{dx} \big[ \ln y \big]= \frac{1}{2} \cdot \frac{d}{dx} \Bigg[ \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg) \Bigg]$$

$$\frac{1}{y} \cdot \frac{dy}{dx}= \frac{1}{2} \cdot \frac{d}{dx} \Bigg[ \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg) \Bigg]$$

$$\frac{dy}{dx}= \frac{y}{2} \cdot \frac{d}{dx} \Bigg[ \ln \bigg( \frac{x(x+2)}{(2x+1)(3x+2)} \bigg) \Bigg]$$

From here, utilize the chain rule as well as the quotient rule.

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