Q.Find the equations of a line tangent to $y = x^3 - 2x^2 + x - 3$ at the point $x = 1$.

Derivatives of Algebraic Functions Using the Chain Rule

The Intuition: Why Do We Need This?

You already know how to differentiate simple functions like $x^2$ or $\sin x$. But what about something like $(3x^2 + 5)^7$? You could expand it — but that would be a nightmare. Or what about $\sqrt{1 - x^2}$? Expanding is impossible.

The chain rule solves this by letting you differentiate functions inside other functions — compositions. Think of it like peeling an onion: you differentiate the outer layer, then multiply by the derivative of the inner layer.

The Core Idea: A Real-World Analogy

Imagine a machine that first doubles a number, then cubes the result. If you input $x$, the machine does:

1. Inner function: $u = 2x$ (doubling)
2. Outer function: $y = u^3$ (cubing)

Now, how fast does the final output change when you change $x$? It's not just the rate of the outer function, nor just the inner. It's the product of the two rates:

• The outer function changes at rate $\frac{dy}{du} = 3u^2$ (with respect to its own input $u$).
• The inner function changes at rate $\frac{du}{dx} = 2$ (with respect to $x$).

So the overall rate is:

That's the chain rule: differentiate the outside, keep the inside the same, then multiply by the derivative of the inside.

The Precise Statement

If $y = f(u)$ and $u = g(x)$, then $y$ is a function of $x$ through $u$. The derivative is:

In function notation, if $h(x) = f(g(x))$, then:

The key: you evaluate the derivative of the outer function at the inner function (not at $x$), then multiply by the derivative of the inner function at $x$.

Step-by-Step Examples

Example 1: Differentiate $y = (3x^2 + 5)^7$

• Outer function: $(\cdot)^7$, derivative: $7(\cdot)^6$
• Inner function: $u = 3x^2 + 5$, derivative: $6x$

Apply the rule:

Example 2: Differentiate $y = \sqrt{1 - x^2}$

Rewrite: $y = (1 - x^2)^{1/2}$

• Outer: $(\cdot)^{1/2}$, derivative: $\frac{1}{2}(\cdot)^{-1/2}$
• Inner: $u = 1 - x^2$, derivative: $-2x$

Apply:

Why It Works (Briefly) …