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The Chain Rule - Problem 4 4,369 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Recall that a composite function f(g(x)) is a function that has another function on the "inside." When taking the derivative of a function like this, we use the chain rule. The chain rule states that you first take the derivative of the "outside" function, then multiply it by the derivative of the "inside function." So for a function h(x)=f(g(x)), its derivative would be h'(x)=f'(g(x))*g'(x).

To determine which function is the inside function, look to see which function is "contained" within another function. For example, for exponential functions, look at the power to which e is raised. For logarithmic functions, it will be what is within the logarithm brackets.

Remember that the chain rule can be used with all other rules of differentiation learned so far; this includes rules for deriving exponents, logarithms, the product rule, the quotient rule, etc.

Let’s take a look at another problem. This one’s kind of interesting because we can differentiate the function h(x) equals 1 over x² plus 1 in two ways. You might see that this is a quotient. You could actually use the quotient rule, but you can also use the chain rule. It’s always nice when you have options. It gives you a little more versatility.

Let’s see how the quotient rule would work on this. The derivative would be, and remember you have to identify the low function and the high function. It's low d high, so x² plus 1 times the derivative of the numerator, which is just zero, minus high d low. 1 times the derivative of x² plus 1, which is 2x. Over the square of what’s below.

That’s (x² plus 1)². This first term just disappears. I'll get -2x over (x² plus 1)². So that’s the derivative using the quotient rule.

Let’s see what it looks like using the chain rule. Now using the chain rule, you need to identify an inside and outside function. Furthermore, let me rewrite this in a slightly different form. I’m going to put x² plus 1 inside the function, x to the -1 because that’s the same as the reciprocal function.

H(x) is the same as this. x² plus 1, the quantity to the -1 power. You can see that the outside function is x to the -1. The inside function is x² plus 1. Let’s use the chain rule on that. H'(x) is going to be, first the derivative of the outside function and using the power rule, we pull the minus 1 in front. We get minus 1 times parenthesis, I’ll write the x² plus 1 later, but what does the exponent become? 1 less. -1 minus 1 is -2. So now I’ll put in the x² plus 1.

You still have to multiply by the derivative of the inside function and that’s 2x. What do we have here? We have minus 1 times 2x, -2x. And this x² plus 1 to the -2, that just means 1 over (x² plus 1)². We get the same exact answer and it’s about the same amount of work. Always nice to have two ways of solving a problem when you can. H(x) has derivative -2x over x² plus 1 to the quantity squared.

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