The Chain Rule in Calculus

AnnaWhen you hear “chain rule,” what do you think it’s helping us do?

MarcoDifferentiate a function that has another function tucked inside it. The key move is to differentiate the outer function, keep its input unchanged, and then multiply by the derivative of that input.

AnnaSo how do we tell which part is inner and which part is outer?

MarcoTake y equals three x plus two, all raised to the fifth power. The inner expression is three x plus two. The outer operation raises its input to the fifth power. We can call the inner expression u, so y equals u to the fifth. That makes the layers easier to see.

AnnaAnd once we’ve named those layers, what’s the rule?

MarcoIf y depends on u, and u depends on x, then the derivative of y with respect to x is the derivative of y with respect to u, times the derivative of u with respect to x. In the notation y equals f of g of x, it’s f prime of g of x, times g prime of x.

AnnaLet’s use that on our fifth-power example. What happens first?

MarcoDifferentiate the outer power and leave the inside alone. That gives five times three x plus two, all to the fourth power. The inside, three x plus two, has derivative three. Multiply by three, and we get fifteen times three x plus two, all to the fourth power.

AnnaDoes the same idea work if the outside isn’t a whole-number power?

MarcoYes. Take the square root of two x plus one. We can write the root as the quantity to the power one-half. Differentiating that outer power gives one-half times two x plus one to the negative one-half. Then we multiply by the inner derivative, which is two. The result is one over the square root of two x plus one. This applies where the original function is defined and differentiable, so x must be greater than negative one-half.

AnnaWhat about a trigonometric function? I can imagine someone forgetting there’s still an inner expression.

MarcoExactly. For the sine of x squared, the outer function is sine, and its derivative is cosine of the same input. So we get cosine of x squared, then multiply by the derivative of x squared, which is two x. The answer is two x times cosine of x squared.

AnnaAnd if there are several layers, do we just keep going?

MarcoOne layer at a time. For the sine of x squared plus one, all cubed, start with the cube. That gives three times the sine of x squared plus one, squared. Then differentiate the sine layer: that gives cosine of x squared plus one. Finally, the inside, x squared plus one, contributes two x. Multiply everything, and we get six x times the sine of x squared plus one, squared, times the cosine of x squared plus one.

AnnaSo what should we check before we call an answer finished?

MarcoCheck that every changing inner layer contributed its derivative. Don’t stop after differentiating the outside, and don’t simplify the inside too early. Also, the derivative of sine of u is cosine of u, not negative cosine of u. Negative sine of u is the derivative of cosine of u.