Value Chain Functions

Value chain Rule:

The derivative of the composition of the functions  which is computed as a product of their derivatives. To be precise:
Let f(x) and g(x) be two functions and let (g. f) (x) = g (f(x)) be their composition.

Value Chain rule 1:

If h(x) = g(f(x)), then h'(x) = g'(f(x))f'(x)

Note: In computing the composition g(f(x)), we apply f to x first, and then we apply g to f(x). According to the chain rule, when doing the derivative, we proceed in the opposite order: g is done first, and then f. One more thing, the f(x) part in g'f(x)) states that, while doing the derivative of g, we do not change f(x)(the f(x) term which is usually called the “inside”).

Sometimes we think of the composition in the following way:

y= g(u) and u =f(x) (i.e., y depends io u, and u depends on x). In that case, y depends on x and its derivative is,

Value Chain Rule 2:

If y= g(u) and u=f(x), then dy/dx = dy/du*du/dx

Examples of value chain function:

Value Chain Rule - Example1:

Compute the derivatives of the following functions.

(a)  Y= (x2+1)14

(b) Y = `sqrt (sin x +1)`

(c)  Y = `1/(e^x +2)`

(d) Y = sin(x2 + 1)

(e)  Y = cos(sec x)

( f )   Y = (sin x)2 + sin (x2).

Solution:

(a)  We start by computing the derivative of the power of 14:

Y' = 14(x2+1)13(x2+1)`

= 14(x2+1)132x

= 28(x2+1)13

(b) Writing y = `(sin x+1)^(-1/2)` , we get

Y` =`1/2` `(sinx + 1)^(-1/2)` (sinx+1)`

=  `1/2`  (sinx + 1)-1/2 cos x

=  `1/2` cos x(sin x +1) -1/2.

(c)  Y = (ex+2)-1.thus,

Y` = (-1)(ex+2)-2(ex+2)`

= -ex(ex+2)-2

= -ex/(ex+2)2.

(d) We start by computing the derivative if sin:

Y` = cos (x2+1)(x2+1)`

= 2x cos(x2+1)

(e)  Y` =-sin(sec x) (sec x)`

= - sin (sec x ) sec x tan x.

( f )   We have to be careful about the order:

Y` = 2(sin x)1(sin x)` + cos(x2)(x2)`

= 2sin x cos x +2x cos(x2)

= sin 2x +2x cos(x2)

In simplifying, we used the formula 2 sin x cos c  = sin 2x.

Value chain Rule - Example 2:

Find `dy/dx` for the following functions,

(a)  Y = 4u2 – 3u +2, u = ex+2e2x

(b) Y  = ln u, u=sin x + cos x

Solution:

(a)  By the chain Rule ,

`dy/dx = (dy/(du)) . ((du)/dx)`

= (8u -3)(ex + 4e2x)

= 8((ex + 2e2x) – 3) (ex + 4e2x).

(b) As in (a),

`dy/dx = dy/(du) . (du)/dx`   = `1/u(cos x - sin x) =(cos x -sin x)/(sin x + cos x)`