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Differentiate $\log_3(7x+2)$

I used the chain rule for this equation, making $u=7x+2$ and $g=\log_3u$. I then calculated $u'$ to be $7$ and $g'$ to be $\frac{1}{(7x+2)\cdot \ln3}$. Now all thats left is to multiply $u'$ and $g'$. $$f'(x)=\frac{7}{(7x+2)\cdot \ln3}$$

In my book it says that this is not correct. Any ideas or hints?

TonyK
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Pablo
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    You should always use just enough parentheses to make sure that groupings are clear, and never too few so as to let it be unclear. $\log_37x+2$ is unclear whether you mean $\log_3(7x)+2$, $\log_3(7x+2)$, $(\log_3(7))x+2$ or something else entirely. Without looking at your attempt and seeing your choice of $u$, I would have initially assumed it was the first. – JMoravitz Aug 26 '18 at 15:49
  • Indeed, $\log_3 7x+2$ means (unambiguously) $(\log_3(7x)) + 2$. I have edited the post accordingly. – TonyK Aug 26 '18 at 15:50
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    @JMoravitz or you could have looked at the mathjax code :) – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 26 '18 at 15:50
  • "In my book it says that this is not correct..." There are sometimes many different ways you can write the same thing. What does your book say the answer is? – JMoravitz Aug 26 '18 at 15:51
  • @TheSimpliFire it should be up to the original poster to check to make sure that the MathJax was parsed correctly and outputted the correct intended equation (preferably before pressing the post/submit button), it shouldn't ever be the situation that in order to read a question correctly we need to look at the MathJax code instead of what MathJax actually produced. – JMoravitz Aug 26 '18 at 15:54

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You are correct.

We have $$\log_3(7x+2)=\frac{\ln(7x+2)}{\ln3}$$ and by the Chain Rule, $$\frac d{dx}\ln(7x+2)=\frac1{7x+2}\cdot7$$ so you get that the derivative is $$\frac7{(7x+2)\ln3}$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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