how is chain rule applied to this equation

Question

I came across this in a DE book. It says Observe from the chain rule that

$$\frac{x’}{x} = \frac{d}{dt} \ln x$$

I am trying to understand how to get this equality. I do get that

$$\frac{d}{dx} \ln x =\frac{1}{x}$$

However, its unclear to me how the chain rule is used here to get the $\frac{d}{dt}$ on the RHS.

Answer 1

The chain rule is as follows:

Suppose that $f$ is continuous on $[a, b]$. Suppose that $f$ is differentiable at the point $t \in [a, b]$. Suppose that $g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(t)$. Define the function $h$ by $$ h(s) = g(f(s)) \qquad (a \leq s \leq b). $$ Then $h$ is differentiable at the point $t$, and $$ h'(t) = g'(f(t)) f'(t). $$

In other words, the chain rule states that $$ {\mathrm{d}y \over \mathrm{d}t} = {\mathrm{d}y \over \mathrm{d}x} \cdot {\mathrm{d}x \over \mathrm{d}t}. $$

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Starting with the right-hand side, one yields $$ {\mathrm{d}(\ln {x}) \over \mathrm{d}t} = {\mathrm{d}(\ln {x}) \over \mathrm{d}x} \cdot {\mathrm{d}x \over \mathrm{d}t} = \frac{1}{x} \cdot x' = \frac{x'}{x}. $$

It is not correct that $$ {\mathrm{d}u \over \mathrm{d}t} = \ln {x(t)}, $$ in which $u = \frac{1}{x(t)}$.

Answer 2

Let us start with the RHS of the equation, and expand using the chain rule:

$$\frac{\mathrm{d}}{\mathrm{d}t}\ln x = \frac{\mathrm{d}\ln x}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{x} x' = \frac{x'}{x}\,\text{.}$$