Confused about substitution step.

[dim lagoon]

Yanek

[mint rover]

in the book im reading the substitution is $\ln x = u^3$ $\frac{dx}{x} = 3u^2 du$ but I don't understand why can you derive the first function $\ln x$ when the derivative is over $du$?

[dim lagoon]

Yanek

[junior copper]

In the substitution you let $[u(x)]^3=ln(x)$. By implicit differentiation you can differentiate both sides by the variable $x$ so $\frac{d}{dx}{[u(x)]^3}=\frac{d}{dx}ln(x)$. The chain rule gives $\frac{d}{dx}{[u(x)]^3}=3 \frac{du}{dx}[u(x)]^2$, and $\frac{d}{dx}ln(x)=\frac{1}{x}$. So $\frac{1}{x} = 3 \frac{du}{dx}[u(x)]^2 \iff \frac{dx}{x} = 3 [u(x)]^2 \cdot du$.

[dim lagoon]

Crystopher