What is the derivative of log y with respect to x?

Question

I was solving the following sum: We started off with the first equation and then he took log on both sides to simplify. Everything was fine here but the next step's LHS confused me, he said he used chain rule to simplify logy, he took y equal to u, and so he got 1/y dy/dx, but I just cant see how this is possible, i know and understand chain rule, but this just seems wrong. Someone please help. Even the textbook does the same thing.

Answer 1

Bear in mind there's no such thing as "the derivative"; it has to be the derivative with respect to something specific. In this case, we differentiate both sides with respect to $x$ (you can't very well do something different to each side). If $y$ is a function of $x$ only, the effect on the left-hand side is to differentiate with respect to $y$, then multiply by $dy/dx$.

Answer 2

If you have a function $f$ of the variable $x$ which admits a derivative, if $\log f$ is define, then it admits a derivative too and \begin{align} \log(f(x)) ' = f'(x) \times \log' (f(x)) = f'(x) \times \frac{1}{f(x)} \end{align} (this is the chain rule)

Answer 3

$\log y$ is a function of $y$, and $y$ is a function of $x$. Then by chain rule $\displaystyle \frac{d}{dx}(\log y)=\frac{d}{dy}(\log y) \times \frac{d}{dx}(y)=\frac{1}{y}\times \frac{dy}{dx}.$

Answer 4

Define $L(x) = \log f(x)$. Take the derivative, $L'(x) = \frac{d L(x)}{dx} = \frac{f'(x)}{f(x)}$ by definition of logarithm derivative. Now rewrite this as $$ f'(x) = f(x) L'(x) $$

The only challenge is to take the derivative of the expression after taking the logarithm, but this is easier than the original expression, because instead of taking derivatives of a fraction you take derivatives of differences (since $\log \frac{a}{b} = \log a - \log b)$.