Derivative a logaritmic function

Question

Let's take $ f = \ln(x) $. The derivative is $ f' = 1/x$.

However $g = \ln(50x) $ has the same derivative $f' = g'$. How come?

If I where going to derivative $g$ I would substitute $x$ for $t$:

$g = \ln(50x) = [50x = t] = \ln(t)$ The derivative should then be $g'=1/t = 1/50x$.

What am I missing?

Answer 1

It might be easier if you realise that $\ln(50x)=\ln(50)+\ln(x)$. Then it becomes obvious why it has the same derivative as the natural logarithm.

Answer 2

It would be this:

Let $g(x)=ln(50x)$, I think better not to let $t=50x$ for the purpose of chain rule.

Now for the derivative of $g(x)$ which is $g'(x)$ we have:

$\frac{1}{50x}\cdot 50=\frac{1}{x}$.

Note: If $g(x)=(f o h)(x)=f(h(x))$, then $g'(x)=f'(h(x))\cdot h'(x)$. Here $f(x)=\frac{1}{x}$ and $h(x)=50x$.

Answer 3

In your notation, $g(x) = f(t(x)) = \ln(50x)$
By the chain rule: $g'(x) = f'(t(x))t'(x) = \frac{1}{50x}\cdot 50 = 1/x$