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I am having difficulties with a problem that asks us to take the derivative of a function that has many different functions inside of it.

The question is as follows:

Let $$f(x) = \ln\left(\frac{(x+1)^4(3x-1)^2}{\sqrt{x^2 + 7}}\right)$$

Find $\frac{df}{dx}$.

There is also a hint given suggesting to simplify this equation first before taking the derivative.

What I have tried so far is using the logarithm laws to rewrite the equation out and then take the derivative. However each time I do this I come up with a different answer than what is given.

I have also tried expanding out the brackets in the numerator and simplifying it as much as possible then applying the logarithm rules but still I am not able to produce the correct answer.

My question is and would like to discuss if this is possible; what is the best method to approach this type of question and how should I simplify this equation into a form that is easier to take the derivative?

I have not had much practice at working with complicated expressions such as this one as I am unsure how I should be applying the chain rule or quotient rule or even how best to simplify this.

Thank you.

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$$f(x) = \ln\left(\frac{(x+1)^4(3x-1)^2}{\sqrt{x^2 + 7}}\right)=4\ln (x+1)+2\ln(3x-1)-\frac{1}{2}\ln(x^2 + 7)$$ then $$f'(x)=4\frac{1}{x+1}+2\frac{3}{3x-1}-\frac{1}{2}\frac{2x}{x^2+7}=\frac{4}{x+1}+\frac{6}{3x-1}-\frac{x}{x^2+7}$$

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    $\begingroup$ The third denominator is $x^2 + 7$, you have made a small error there. $\endgroup$ – Shailesh Oct 5 '15 at 10:50
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    $\begingroup$ Thank you. Once I saw the simplification which you had given from the hint I was able to follow through with the question.Thank you. I was a bit unsure about my third term but the other comment has cleared everything up thank you :) $\endgroup$ – user241451 Oct 5 '15 at 10:52
  • $\begingroup$ @Shailesh: Oops! I didn't even notice that typo when I was improving the LaTeX. $\endgroup$ – PM 2Ring Oct 5 '15 at 10:53
  • $\begingroup$ NP. You still have time to edit it. Though the answer has already got 2 upvotes !! $\endgroup$ – Shailesh Oct 5 '15 at 10:56
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    $\begingroup$ @Michael, you are welcome, thanks for editing guys :) $\endgroup$ – R.N Oct 5 '15 at 11:07

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