How the Chain Rule Actually Works (And What Khan Academy Gets Right)
The chain rule is one of those calculus tools that feels like it should be harder than it actually is. On paper it looks simple enough — you have a function inside another function and you need the derivative. In practice, people mess it up because they treat it like a formula to memorize instead of a process to follow step by step. Khan Academy has one of the more straightforward explanations of the chain rule online, and their video approach of isolating the inner and outer functions before differentiating is genuinely useful. Here is how the method works when you actually sit down and use it. You are given something like the derivative of sin(x²). The first step is identifying what is inside and what is outside. The inner function is x squared. The outer function wraps around it — in this case, sine. You differentiate the outer function while keeping the inner function exactly where it is, then you multiply by the derivative of the inner function. That is it. The formula is f prime of g of x times g prime of x, but writing it out as a process helps more than memorizing that notation. When I first tried teaching this to students, I kept seeing the same mistake. People would differentiate the outer function and then just drop the answer without multiplying by the inner derivative. It happens constantly. The reason is that the chain rule has two multiplication steps baked into it, and students often complete the first one and forget the second exists. Khan Academy's videos address this by showing the full unsimplified expression before combining anything, which forces you to see both pieces.
Here is a harder example that trips people up. Take the derivative of e to the power of 3x plus 1. The inner function is 3x plus 1. The outer function is the exponential. You take the derivative of the exponential, which is still the exponential, keep the inner function untouched, and then multiply by 3, the derivative of the inner part. The answer is 3e to the power of 3x plus 1. A lot of students write just e to the power of 3x plus 1 and stop there because the outer derivative looks identical to the original outer function. That is a real trap with exponentials and logarithms — the derivative of the outside doesn't visibly change, so you second-guess whether you missed a step. You didn't. The missing piece is always the inner derivative. I ran into a specific edge case recently that most introductory resources skip over entirely. A student sent me a problem involving a nested chain rule where the function was something like the square root of the natural log of x to the fifth power. That is three layers: the fifth power is inside the natural log, which is inside the square root. Khan Academy's standard chain rule tutorial covers two layers well, but three layers is where things get messy. The workaround I used was to label each layer explicitly before differentiating anything. I wrote out layer one as x to the fifth, layer two as the natural log of layer one, and layer three as the square root of layer two. Then I differentiated from the outside in, writing each derivative as a separate multiplication factor. The result had three multiplicative terms instead of two. Without that labeling step, it is very easy to drop one of the derivatives and get an answer that is off by a factor of 5x to the fourth or something equally hard to catch. Another thing that people miss is when the chain rule is not actually needed. If you see something like the derivative of sin squared of x, that looks like it requires the chain rule. It does. But if you see the derivative of sin of x squared, that is also a chain rule problem, and the placement of the exponent matters enormously for how you approach it. sin squared x means sin of x raised to the second power, which is (sin x)². The outer function is the square and the inner function is sin x. But sin of x squared means sin of x², where the outer function is sine and the inner function is x². These produce completely different answers. Students routinely conflate the two notations because they look similar on paper.
There is also a scenario where the chain rule applies in reverse and becomes part of u-substitution in integration. Khan Academy covers this in their integration videos, but the connection is not always obvious when you first learn the chain rule. Recognizing that integration by substitution is essentially the chain rule working backward will save you a lot of time later on. When you see an integral with a composite function and the derivative of the inner function sitting alongside it, that is your signal that u-substitution is the path forward. This pattern recognition tends to come only after you have differentiated enough chain rule problems that the structure becomes familiar. The main limitation of relying solely on Khan Academy for this topic is that their examples tend to stay on the clean side. Real exam questions and higher-level calculus problems will nest functions in ways that require simplification before you even start differentiating. Sometimes you need to expand or rewrite an expression using algebra or trigonometric identities before the chain rule becomes apparent. Khan Academy occasionally skips this preprocessing step, which means students can feel lost when they encounter a problem that does not immediately reveal its inner and outer functions. The workaround is to practice rewriting expressions before differentiating. Take (2x plus 1) to the third power and expand it first, then differentiate term by term. Compare that answer to what you get using the chain rule directly. They match. This exercise builds confidence that the chain rule is not a separate method but a shortcut that works consistently. If you want to work through the Khan Academy Chain Rule material yourself, their calculus section is freely available at Khan Academy. The specific videos walk through the formula, show several worked examples, and include practice problems with step-by-step hints. It is not the most polished resource on the internet, but it is one of the clearest and it is completely free. For anything beyond the basic two-layer problems, you will want to supplement it with additional practice sets, preferably ones that include the three-layer nesting cases and the algebra-preprocessing scenarios I mentioned earlier.
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