Remember that the chain rule is used to find the derivatives of composite functions. First, determine which function is on the "inside" and which function is on the "outside." The chain rule will be the derivative of the "outside" function multiplied by the derivative of the "inside" function. The chain rule can be used along with any other differentiating rule learned thus far, such as the power rule and the product rule. So, when finding the derivative of some product involving a composite function, use the chain rule to find the derivative of the composite part, and then use the product rule as you normally would: the derivative of h(x)=f(x)g(x) is h'(x)=f(x)g'(x)+f'(x)g(x).
We’re using a special case of the chain rule that I call the general power rule. Here’s a problem that we can use it on. Differentiate y equals x² times the square root of x² minus 9. Whenever I’m differentiating a function that involves the square root I usually rewrite it as rising to the ½ power.
This is going to be x² times x² minus 9 to the ½ power. Because I’m going to use the general power rule on this piece of the function. But notice this part of the function is also a product. So I’m going to have to use the product rule. Just making sure that you see this; this left piece is one of the factors in the product and this right piece is another. I’ll use the general power rule on this piece but I don’t need the general power I just need the regular power rule for this guy.
Let’s differentiate. We have dy/dx equals, now I’m using the product rule here; the first times the derivative of the second. The first was x² and then we want the derivative of, and I’m going to, I’m not going to actually do the derivative I just want to represent it by using this notation. The derivative of x² minus 9 to the ½ plus the second x² minus 9 to the ½, times the derivative of the first.
This step is just the product rule. I haven’t actually done any actual derivatives yet. Next, x² times, and here’s where I need the general power rule. The ½ comes in front and I have (x² minus 9) to the -1/2. Replace this exponent with 1 less. So -1/2 times the derivative of the inside function and that’s the derivative of this. It’s going to give me 2x. Plus and now I’m on this term, x² minus 9 to the ½ times the derivative of x², 2x.
Let’s see how this combines. We have a ½ and a 2 which shall cancel. So these cancel. I have an x and an x ², that makes x ³, and this x² minus 9 to the ½, it’s like 1 over x² minus 9 to the +½. So that becomes the square root of x² minus 9 in the denominator. And over here, I’ve got 2x times the square root of x² minus 9. And that’s my derivative of the function y equals x² times root x² minus 9.