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:

h(x) = (3x 2 + 5x 1) 4 : outer function f is ( thing ) 4 , inner is thing = g(x) = ( 3x 2 + 5x 1).
f '(x) = 4 ( thing ) 3 , and g '(x) = 6x + 5
so h '(x) = f '[ g(x) ]· g '(x) = 4 ( 3x 2 + 5x 1) 3 ( 6x + 5)
Notice that when we find f '(x) -- on the outside layer -- we leave g(x) unchanged.
We're dealing strictly with the function f (x) -- once that's done -- we find g '(x) and multiply.
So, if h (x) = ( something ) n , h '(x) = n ( something ) n - 1 (something ) '

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

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 4x² + 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) 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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