Product & chain rule

The power rule and the sum rule take you a long way with polynomials. But $f(x) = x^2 \cdot \sin x$ is a product of two different functions, and $f(x) = (3x + 1)^5$ is one function sitting inside another. There is exactly one rule for each case, and once you know them you can differentiate almost anything you meet.

Why the easy way fails

The obvious idea would be to differentiate both parts separately and multiply. That is wrong, and you can see it in an example where you already know the answer. Take $f(x) = x \cdot x = x^2$. The correct derivative is $2x$. Differentiating separately and multiplying would give $1 \cdot 1 = 1$, which is plainly not right.

The product rule

For a product $u \cdot v$ you differentiate one factor and leave the other alone, then the other way round, and add the two. It can be remembered as a phrase: first differentiated times second left alone, plus first left alone times second differentiated.

The check on $f(x) = x \cdot x$ now works: $u = x$, $v = x$, so $u' = 1$ and $v' = 1$. Substituting gives $1 \cdot x + x \cdot 1 = 2x$, exactly the familiar derivative of $x^2$. A real example: $f(x) = x^2 \cdot \sin x$ has the derivative $2x \cdot \sin x + x^2 \cdot \cos x$.

The chain rule

In $f(x) = (3x + 1)^5$ one function sits inside another. Inside is $3x + 1$, outside is the fifth power. You could expand the bracket, but that would be six terms of work for a task that fits on one line. The chain rule says: differentiate the outer function, leave the inner one untouched while doing so, and multiply by the derivative of the inner function at the end.

For $f(x) = (3x + 1)^5$ that means: the outer derivative is $5 \cdot (3x + 1)^4$, the inner derivative of $3x + 1$ is 3. Together $f'(x) = 15 \cdot (3x + 1)^4$. The factor 3 at the end is the step people forget most often, which is why it gets a name of its own.