Chain Rule and u-Substitution: Example with y = \ln(x^4 + 2)

Infinitely many infinitely thin rectangles: a way people visualize area under a curve. The idea leads to limits and integration concepts.
The statement Infinity times zero is undefined: \infty \cdot 0 is not a well-defined number; a formal treatment uses limits instead.
The goal is to formalize how to handle limits and inner-outer function relationships when differentiating or integrating.
Acknowledgement of a transition to working with x and y explicitly, and the introduction of an intermediate variable (u) to simplify the process.

There are many notations for derivatives and several ways to write the chain rule; the common intuition is that the chain rule looks like fractional cancellation, but it is not cancelling fractions—it's a product of derivatives.
The chain rule can be written as:
- \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
The idea is to introduce an intermediate variable u to simplify the derivative when y is a function of u, which is itself a function of x.
In practice, you choose u to be a function of x that makes the expression easier to differentiate.

Define the inner substitution: u = x^4 + 2
Then define the outer function: y = \ln(u)
Compute the necessary derivatives:
- \frac{dy}{du} = \frac{1}{u}
- \frac{du}{dx} = \frac{d}{dx}(x^4 + 2) = 4x^3
Apply the chain rule:
- \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 4x^3
- Substituting back for u yields:
  \frac{dy}{dx} = \frac{4x^3}{x^4 + 2}
Note on notation:
- The expression "1 over u times 4x^3" comes from treating dy/du and du/dx as separate derivatives.
- The appearance of a fraction in the chain rule notation should be understood as a product of derivatives, not literal cancellation of factors.
Conceptual takeaway: this approach uses the inner function u to organize derivatives so that dy/dx can be written in terms of x (or in terms of u after substitution).

The inner function is often denoted f or, here, u, with a derivative that you try to “pull into” the numerator.
If you view the inner function as f(x) = x^4 + 2, then f'(x) = 4x^3, which sits in the numerator when you apply the chain rule in the form dy/dx = (dy/df) (df/dx) with y = something of f.
The speaker describes the sense of being “greedy” about putting the inner derivative in the numerator to make the structure clear and to set up substitution nicely.
This perspective aligns with the idea of choosing substitutions that render the integrand a simple function of the new variable (u) instead of a messy composition in x.

Although the example is a differentiation scenario, the same substitution mindset applies to integration (u-substitution): choose a u that is the inner function so that du appears in the integrand.
In integration, if you have an integral involving a composite function, you try to spot a part whose derivative also appears in the integrand, enabling a clean substitution to simplify the integral.
A common related result (for intuition) is: if you encounter an integrand of the form \frac{f'(x)}{f(x)}, then the antiderivative is \ln|f(x)| + C, which mirrors the chain-rule structure from differentiation.

When using substitution (u-substitution) in differentiation or integration:
- Identify an inner function u = g(x) that simplifies the expression.
- Compute du/dx and relate dy/dx (or the integrand) to dy/du and du/dx.
- Remember that the chain-rule notation resembles cancellation but is actually a product of derivatives; treat it as such.
In practice, the choice of u is strategic: pick the inner function so that the derivative du/dx appears naturally in front of the rest of the expression.
The approach is useful in various scenarios; practice with different forms to build fluency.

Inner substitution: u = x^4 + 2
Outer function: y = \ln(u)
Derivatives:
- \frac{\mathrm{d}y}{\mathrm{d}u} = \frac{1}{u}
- \frac{\mathrm{d}u}{\mathrm{d}x} = 4x^3
Chain-rule result: \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{u} \cdot 4x^3 = \frac{4x^3}{x^4 + 2}

The instructor mentions planning to show the next clicker question, indicating this is part of a broader set of practice problems on substitution or chain-rule applications.
Mathematical symbols in this note use LaTeX formatting where helpful to display derivatives and substitutions clearly. For reference:
- u = x^4 + 2
- y = \ln(u)
- \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 4x^3 = \frac{4x^3}{x^4 + 2}