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?
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 $$\frac7{(7x+2)\ln3}$$