For a question in my textbook:
Differentiate $\log(2x)$
The differentiation rule for logarithm is $1/x \ln b$, where $b$ is the base. So my answer was $1/(2x) \ln 10$, but the answer my textbook gave in the back was $1/x \ln 10$. When I substituted $7236x$ for $2x$ in the original equation, the derivative of $\log(7236x)$ is still $1/x \ln b$, not $1/7236x \ln b$. Where did I go wrong? Why is it the way it is?
On
The most direct way to attack the problem is to notice that $$\log(cx)=\log(c)+\log(x)$$ and then obviously, on the right side, we have a constant plus $\log(x)$ - and taking the derivative eliminates the constant term. That aside, you used the chain rule incorrectly in your solution, which is why it gets the wrong answer. If we differentiate $$f(2x)$$ we get $$2f'(2x)$$ but what you've written is just $$f'(2x)$$ - that is, you forgot to multiply by the derivative of the inner function $2x$. If $f(x)=\log(x)$, then you correctly wrote out $f'(2x)=\frac{1}{2x}\ln(10)$, but you forgot the factor of two - yielding $$2\cdot \frac{1}{2x}\ln(10)=\frac{1}x\ln(10)$$ which is the correct answer.
You need to use the chain rule. For $f(x)=\ln ax$, $$f'(x)=\frac 1{ax}a=\frac 1x.$$