At first when we began to learn the chain rule, I was kind of confused. I understood the concept, and how to use it. However, I was not completely sure if I would know when I should use it or which problems it should most definitely be applied to. Then, I realized that we had been using the chain rule all along without really paying any attention to it. When we would find the derivative of problems using the constant rule, power rule, product rule, and quotient rule, we were still following the principles of the chain rule. For example:
f(x) = 4x^5 - 2x^4 + x^2 + 8
Using the product rule (bring the exponents to the front, and subtract them by one) the derivative would look like:
f'(x) = 20x^4 - 8x^3 + 2x
This was simple to do, and the chain rule was unnecessary. It becomes necessary to use the chain rule, however, when we start to look at a composition of functions equation because there are inside and outside functions present.
With the following equation, the chain rule needs to be used:
f(x) = 3(x^2 +5)^2
There are two approaches to the chain rule. The first would be to use reasoning to find the derivative:
The derivative of the outside of the function is: 6(x^2 + 5)
The derivative of the inside of the function is: 2x
I am left with the derivative of the entire function after multiplying the derivatives of the inside and outside functions:
f'(x) = [6(x^2 + 5)] [2x] --------> f'(x) = 12x(x^2 + 5)
The second approach would be to call the inside function (u). I do this because I believe that the further we get into finding derivatives of more complicated functions or even finding integrals of derivative functions, they are much easier to accomplish with the use of (u).
f(x) = 3(x^2 +5)^2 can be re-written as 3(u)^2, where u = x^2 + 5
The derivative of the outside of the function is: 6(u)
The derivative of (u) is: 2x
Now, I can substitute for (u) when multiplying the derivatives of 6(u) and 2x, and I will still end up with:
f'(x) = 12x(x^2 + 5)
Some of the equations that really confuse me are the ones that use square roots, cube roots, etc., and also some of the trig. functions. Sometimes I get lost when doing composition of functions equations that have two inside functions and an outside function with trig. elements in them simply because I have a hard time remembering that whatever is in place of the (x) in the equation, is in the place of (x) in the derivative. For example:
f(x) = tan(sec(4x + 2))^3
First, I have to take the derivative of the outside function: 3tan(sec(4x + 2))^2
Then, I have to find the derivative of the first inside function: sec(4x + 2)tan(4x + 2)
Sometimes I forget that the (4x + 5) applies to the (x)'s in the derivative of sec(x) since they are in the original function.
Then, I have to find the derivative of the final inside function: 4
Now, I have to multiply the three derivatives:
f'(x) = [3tan(sec(4x + 2))^2] [sec(4x + 2)tan(4x + 2)] [4]
And, simplified the answer is:
f'(x) = 12tan[sec(4x + 2)]^2 sec(4x + 2)tan(4x + 2)
Other than equations similar to that I found it somewhat simple to understand once I got the hang of doing the equations!
f(x) = 4x^5 - 2x^4 + x^2 + 8
Using the product rule (bring the exponents to the front, and subtract them by one) the derivative would look like:
f'(x) = 20x^4 - 8x^3 + 2x
This was simple to do, and the chain rule was unnecessary. It becomes necessary to use the chain rule, however, when we start to look at a composition of functions equation because there are inside and outside functions present.
With the following equation, the chain rule needs to be used:
f(x) = 3(x^2 +5)^2
There are two approaches to the chain rule. The first would be to use reasoning to find the derivative:
The derivative of the outside of the function is: 6(x^2 + 5)
The derivative of the inside of the function is: 2x
I am left with the derivative of the entire function after multiplying the derivatives of the inside and outside functions:
f'(x) = [6(x^2 + 5)] [2x] --------> f'(x) = 12x(x^2 + 5)
The second approach would be to call the inside function (u). I do this because I believe that the further we get into finding derivatives of more complicated functions or even finding integrals of derivative functions, they are much easier to accomplish with the use of (u).
f(x) = 3(x^2 +5)^2 can be re-written as 3(u)^2, where u = x^2 + 5
The derivative of the outside of the function is: 6(u)
The derivative of (u) is: 2x
Now, I can substitute for (u) when multiplying the derivatives of 6(u) and 2x, and I will still end up with:
f'(x) = 12x(x^2 + 5)
Some of the equations that really confuse me are the ones that use square roots, cube roots, etc., and also some of the trig. functions. Sometimes I get lost when doing composition of functions equations that have two inside functions and an outside function with trig. elements in them simply because I have a hard time remembering that whatever is in place of the (x) in the equation, is in the place of (x) in the derivative. For example:
f(x) = tan(sec(4x + 2))^3
First, I have to take the derivative of the outside function: 3tan(sec(4x + 2))^2
Then, I have to find the derivative of the first inside function: sec(4x + 2)tan(4x + 2)
Sometimes I forget that the (4x + 5) applies to the (x)'s in the derivative of sec(x) since they are in the original function.
Then, I have to find the derivative of the final inside function: 4
Now, I have to multiply the three derivatives:
f'(x) = [3tan(sec(4x + 2))^2] [sec(4x + 2)tan(4x + 2)] [4]
And, simplified the answer is:
f'(x) = 12tan[sec(4x + 2)]^2 sec(4x + 2)tan(4x + 2)
Other than equations similar to that I found it somewhat simple to understand once I got the hang of doing the equations!