Chain Rule for Derivatives |
Though the Chain Rule causes lots of headaches it is extremely useful. It deals with layers of functions -- one function applied to another, so when we apply it, we pretend to peel the layers like peeling an onion. If we have to differentiate ( x 7 )² , it's no problem to square the binomial and then find the derivative. However, if we have to differentiate ( 3x 7 )27, it could take us a week to expand the polynomial and then differentiate. This is where the chain rule saves the day.
The Chain Rule states:
( f [ g(x) ] )' = f '[ g(x) ]· g '(x)
This says that we take the derivative of the outer function f ( x ) first, then we multiply it by the derivative of the inner function g ( x )
1st: take f ' , the derivative of the outer function -- leave inside unchanged!
2nd: multiply by the derivative of the inner function.
Example:
Examples:
function | derivative |
( x³ + 9x 5 ) 7 . | 7( x³ + 9x 5 ) 6 ( 3x² + 9 ) |
( 4x² 15 ) 1/ 4 . | ¼ ( 4x² 15 ) 3/ 4 ( 8x ) = 2x ( 4x² 15 ) 3/ 4 . |
3(x 5 + 7x 3 + 2x 17) 15 | 45(x 5 + 7x 3 + 2x 17) 14 (5x 4 + 21x 2 + 2) |
product & chain rule (2x + 7) 3 (4x 3) 5 |
3(2x + 7) 2 (2)(4x 3) 5 + 5(4x 3) 4 (4)(2x + 7) 3 |
quotient & chain rule![]() |
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Now let's learn to clean up the mess we get from the last 2 examples -- with 2 rules together.
The generic answer for the 4th example -- with both product and chain rule was:
3(2x + 7) 2 (2)(4x 3) 5 + 5(4x 3) 4 (4)(2x + 7)³
First we clean up the constants -- put them at the left end like this:
6(2x + 7) 2 (4x 3) 5 + 20(4x 3) 4 (2x + 7) 3 .
Now we have 2 terms with + between them. Their common factors are:
2 ..... (2x + 7) 2 ..... and ..... (4x 3)4 .
So we factor these out of the 2 humongous terms to get:
2(2x + 7)² (4x 3)4 [ 3 (4x 3) + 10 (2x + 7) ].
When we multiply out the brackets and collect like terms, we get:
2(2x + 7)² (4x 3)4 [ 32x + 61 ]
To find the zeros of this function, we simply set each factor = 0 and solve. The zeros here are:
x = 7/2 ..... x = ¾ ..... and ..... x = 61/32
Now let's do the same for the last example with quotient and chain rule.
Our answer in this one is:
The common factors for the terms in the numerator are 2( x 5) and ( 4x + 9)². Once we factor these out, we get:
We can cancel the ( 4x + 9)² 's top and bottom, then we multiply out the brackets and collect like terms. Here's what we get:
The zeros of this derivative happen at x = 5 and at x = 39/2.
Notation Issues with Trignometric Functions and Chain Rule
The notation for exponentiated trig functions can be confusing. If y = sin 3 (7x) we're really indicating that y = (sin 7x) 3 however, this notation causes confusion since it looks like the (7x) is cubed also -- when it's not! It is the sine function that is cubed!!
The derivative of y = sin 3 (7x) is y ' = (3 sin 2 (7x))( 7 cos 7x)
3 sin 2 (7x) is the derivative of (sin 7x) 3
7 cos 7x is du (7) times the derivative of sin 7x (cos 7x).
we're using the chain rule on 3 levels of functions:
level 1: (thing cubed) ' becomes 3(thing) 2
level 2: sin (something)' becomes cos (something)
level 3: something (7x)' becomes (something)' (7 -- in front!)
Practice Exercises
1) Differentiate with the Chain Rule and any other applicable rules. Simplify the answers:
a) k(x) = (x 4 x² + 1) 1 ...................... b) H(x) = ½(3x² 1) 4
c) .....................d) k(x) = [ x² + (x² + 9) ½ ] ½
e) f(x) = sin² (4x³).................................f) f(x) = (3x + 2)² (7x 5) 6
g) f(t) = (4t³ + 2t 1) ¾ ........................h)
i) Write the equation of the tangent to the function in (h) at the point where x = 1.
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Solutions
function | derivative |
(a) k(x) = (x 4 x² + 1) 1 | k '(x) = (x 4 x² + 1) 2 (4x³ 2x) = |
b) H(x) = ½(3x² 1) 4 . | H '(x) = 2(3x² 1) 3 (6x) = 12x (3x² 1) 3 |
c) ![]() |
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d) k(x) = [ x² + (x² + 9) ½ ] ½ | k '(x) = ½ [ x² + (x² + 9) ½ ] ½ [2x + x(x² + 9) ½ ] |
e) f(x) = sin² (4x³). | f '(x) = 24x² [sin (4x³)] cos(4x³). |
f) f(x) = (3x + 2)² (7x 5) 6 | f '(x) = 6(3x + 2) (7x 5) 6 + 42 (7x 5)5 (3x + 2)² f '(x) = 6(3x + 2) (7x 5) 5 (28x + 9) |
g) f(t) = (4t³ + 2t 1) ¾ . | f '(t) = ¾ (4t³ + 2t 1) ¼ (12t² + 2) |
h) ![]() |
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