The document discusses the chain rule for derivatives. The chain rule is used when taking the derivative of a function composed of an inner and outer function, such as y=f(g(x)). The chain rule states that the derivative of the outer function is multiplied by the derivative of the inner function: dy/dx = f’(g(x)) * g’(x). Some examples are worked out, such as finding the derivative of y=(2x+5)2 and y=sin(x7), to demonstrate how to apply the chain rule.

1. By
Trevor Bennett

2.  The chain rule basically deals with functions
within a function. There is a strong
relationship between them. If you consider
the links on a chain, one link overlaps the
other.
 If there is a change to either the inside or
outside link, it affect the relationship between
them.

3.  I believe so far that you have covered
derivatives of functions. For example y =f(x)
and dy/dx = f’(x).
 But now considertwo functions f(x) and g(x).
We will loo at a function within a function.
 For instance y = f(g(x)). So then what is y’?
 The chain rule is what is applied!!

4.  If y = f(g(x))
 y’= f ’(g(x)) . g’(x) OR

 Alternatively
 You can let u = g(x) and so it becomes
 y = f(u)
 And then to determine y’, we get
 y’ = dy/dx = dy/du * du/dx.

5.  It can also be written as
 y’ = dy/dx = du/dx * dy/du

6.  A simple example could be
 y = (2x +5)2
 Let u = 2x +5 therefore y =u2
 du/dx = 2 + 0 =2
 dy/du = 2u1 = 2(2x +5)
 Therefore dy/dx = dy/du *du/dx
 = 2*2(2x +5)=4(2x + 5)
 = 8x + 20

7.  y = 5/(3x-5)3
 y’ = 5(3x-5)-3
 y’ =-3*5(3x-5)-4 *(3)
 y’ =-45(3x-5)-4
 y’ =-45/(3x-5)4

8. 
 y = sin(x7)
 let u = x7
 y’ = dy/dx = dy/du *du/dx
 dy/du = cos(u) & du/dx = 7x6
 y’ = cos(x7) *7x6
 y’ =7x6*cos(x7)