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 
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.
.
.
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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