When do you use the chain rule instead of the product rule?

2 Answers
Jan 1, 2017

Answer:

Explanation:

Product Rule:

# d/dx (uv) = u(dv)/dx + (du)/dxv#

The Product Rule is used when the function being differentiated is the product of two functions:

Eg if #y =xe^x# where
Let #u(x)=x, v(x)=e^x => y=u(x) xx v(x)#

Chain Rule

# dy/dx = dy/(du) * (du)/dx #

The Chain Rule is used when the function being differentiated is the composition of two functions:

Eg if #y=e^(2x+2)#
Let #u(x)=e^x, v(x)=2x+2 => y = u(v(x)) = (u@v)(x) #

Jan 1, 2017

Answer:

Determining the appropriate differentiation rule is related to the order of operations.

Explanation:

Look at the expression and ask yourself, "If I plugged in a number for the variable, what is the last operation I would do?"

I'll use #x# for the variable in the examples below.

Keep in mind that expressions that are added (or subtracted) can be handled separately. (We differentiate term by term. Things that are added are called "terms".)

If the last operation is to multiply (or divide) by a constant, the back up to the penultimate operation.
("Penultimate" means "next to last".)

If the last operation on variable quantities is multiplication, use the product rule.

If the last operation on variable quantities is division, use the quotient rule.

If the last operation on variable quantities is applying a function, use the chain rule.

#f(x) = 3(x+4)^5# -- the last thing we do before multiplying by the constant #3# is "raise to the #5th# power" -- use the chain rule.

#g(x) = 7x^3(4x-1)^9# -- the last thing we do before multiplying by a constant is multiply two variable expressions -- use the product rule.

Keep in mind that we can rewrite the expression and that doing so may change the appropriate differentiation rule.

#y = (x+3)^2/x# -- quotient

# = (x+3)^2x^-1# -- product

# = (x^2+6x+9)/x# -- quotient

# = x+6+9x^-1# -- power rule going term-by-term.