Chain Rule Fundamentals

The Chain Rule states that if y = u^n where u is a function of x, then the derivative is y' = n·(u)^$n-1$·$du/dx$. This lets you differentiate complex expressions step by step.

Let's see this in action with examples. For y = $x² - 3x + 1$^5, we identify the "outer function" (raising to power 5) and the "inner function" $x² - 3x + 1$. Applying the chain rule gives us y' = 5$x² - 3x + 1$^4·$2x - 3$.

For expressions with negative exponents like y = $x² - 1$^(-5), we follow the same pattern, getting y' = -10$x² - 1$^(-6)·(2x) which simplifies to y' = -20/$x² - 1$^6.

💡 Think of the Chain Rule as peeling an onion - you differentiate one layer at a time, working from the outside in!

More complex expressions like y = $x² + 1$^3·$x³ - 1$^2 require both the Product Rule and Chain Rule. We break it into parts, differentiate each using the Chain Rule, and then combine using the Product Rule. Similarly, when dealing with quotients like y = $(2x - 1)/(3x + 1)$^3, we apply the Quotient Rule inside the Chain Rule.

Chain Rule with Trigonometric Functions

The Chain Rule works beautifully with trigonometric functions. For example, with y = sin(2x), we recognize that the inner function is 2x and the outer function is sine. The derivative becomes y' = cos(2x)·(2) = 2cos(2x).

When dealing with more complex expressions like y = sin$x^5$, we follow the same pattern: y' = cos$x^5$·$5x^4$ = 5x^4·cos$x^5$. Similarly, for y = (cos 3x)^4, we apply the Chain Rule twice: y' = 4(cos 3x)^3·$-sin(3x)$·3 = -12cos^3(3x)·sin(3x).

🔑 Remember: First identify the "outer" and "inner" functions, then apply the Chain Rule formula y' = (outer function derivative)·(inner function derivative).

The Chain Rule also applies to exponential functions. For y = e^(ax), the derivative is y' = e^(ax)·(ax)'. And for y = e^$-t$, we get y' = e^$-t$·(-1) = -e^$-t$. This demonstrates how the Chain Rule lets you differentiate virtually any composite function you'll encounter in your calculus course.