$$f(x) = \ln \sqrt{4x-3}$$ I'm practicing derivation for the exam and I'm stuck on this task. Could someone help me out in solving this. But when it comes to root, I'm a bit confused. The result is supposed to be $\frac2{4x-3}$.

I know for example if it were $f(x)=\ln(x^5)$, I would know how to solve it and get the result which is $$f'(x) = \frac1{x^5} \cdot 5x^4 = \frac5x$$

  • $\begingroup$ For $4x-3>0,$ $$\ln(4x-3)^{1/2}=\dfrac{\ln(4x-3)}2$$ $\endgroup$ – lab bhattacharjee Apr 27 '15 at 14:23
  • $\begingroup$ Ooooh, that's what I needed okay thanks :P $\endgroup$ – Narraxus Apr 27 '15 at 14:24
  • $\begingroup$ @AlexR, Look at OP's comment $\endgroup$ – lab bhattacharjee Apr 27 '15 at 14:25
  • $\begingroup$ @labbhattacharjee, when $4x-3=0$, $\ln\left((4x-3)^{1/2}\right)$ is undefined. $\endgroup$ – Prasun Biswas Apr 27 '15 at 14:26
  • $\begingroup$ @labbhattacharjee Not sure what OP has to do with this. It was more about readability than about correctness. Nevermind. I'm deleting the comment since it's too late anyway. $\endgroup$ – AlexR Apr 27 '15 at 14:39

If you want to do it without simplifying first, you can apply the chain rule two times: $$f'(x) = \underbrace{\frac1{\sqrt{4x-3}}}_{= \ln'\sqrt{4x-3}} \cdot \underbrace{\frac1{2\sqrt{4x-3}}}_{= \sqrt{\ \ }' (4x-3)} \cdot \underbrace{4}_{= (4x-3)'} = \frac2{4x-3}$$ Using the simplification, it's only one chain rule $$f'(x) = \underbrace{\frac12}_{\text{const.}} \cdot \underbrace{\frac1{4x-3}}_{\ln'(4x-3)} \cdot \underbrace{4}_{(4x-3)'} = \frac2{4x-3}$$


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