Function of a Function Mathematics N5

 

1.5 Chain Rule.

The Chain Rule for Function of Function is as stated.

      =     X                                                                                           =     X     X                                                                                The substitution of u reduces the function of a function problem to the most basic function.   Function of a function problems may be classified as stated in the tables *******.

 

 

 

 

Table 1.4

Category A. f’ (x) ≠ 1
	Basic Function	Substitution u=	Function of a Function
	y		=   X
1.	ex	f(x) = 2x	e2x
2.	Inx=loge x	f(x) =x3	Inx3=loge x3
3.	ax	f(x) = 0.5x	a0.5x
4.	sinx	f(x) = x2+2x	Sin(x2+2x)
5.	cosx	f(x) = Inx	Cos(Inx)
6.	tanx	f(x) = logx	Tan(logx)
7.	Cosecx	f(x) = 3x3+2x2	Cosec(3x3+2x2)
8.	Secx	f(x) = 5x2+4x	Sec(5x2+4x )
9.	cot 50x	f(x) = 50x	cot 50x

Table 1.5

	Function of Function	Substitution u=	Chain Rule Problem in terms of u
	y		=   X
1.	e2x	f(x) = 2x	eu
2.	Inx3=loge x3	f(x) =x3	In u =loge u
3.	a0.5x	f(x) = 0.5x	a0.5x
4.	Sin(x2+2x)	f(x) = x2+2x	Sin u
5.	Cos(Inx)	f(x) = Inx	Cos u
6.	Tan(logx)	f(x) = logx	Tan u
7.	Cosec(3x3+2x2)	f(x) = 3x3+2x2	Cosec u
8.	Sec(5x2+4x )	f(x) = 5x2+4x	Secu
9.	cot 50x	f(x) = 50x	cot u

 

Table 1.6 Function of a function in  terms of power greater than 1 or less than 1.

Category B. Exponent of standard function is less than 1 or greater than 1
	Basic Function	New Exponent	Function of a Function
	y
2.	(ex)1	0.5	e0.5x
3.	(Inx)1= (loge x)1	50	(Inx )50 = (loge x)50
4.	(ax)1	20	a20x
5.	(Sinx)1	-0.3	(Sinx)-0.3
6.	(Cosx)1	7	(Cosx)7
7.	(Tanx)1	18	(Tanx)18
8.	(Cosecx)1	0.6	(Cosecx)0.6
9.	(Secx)1	55	(Secx)55
10.	(cot x)1	13	(cot x)13

 

 

 

1.6 Differentiation Tables in terms of f(x).

The derivation of the formulas in the tables are based on the chain rule as stated previously. It is very fundamental for students to differentiate based on the method of tables and by method of chain rule.

 

Table 1.7

	Differentiation	Tables
	y=	Differential
1.	K[f(x)]n	Kn[f(x)]n-1f’(x)
2.	Kef(x)	Kf’(x)ef(x)
3.	Kaf(x)	Kf’(x)af(x) Ina
4.	KIn f(x)	K[f’(x)]/[f(x)]
5.	KIoga  f(x)	K[f’(x)]/[f(x)Ina]
6.	Ksin f(x)	Kf’(x)cosf(x)
7.	Kcos f(x)	-Kf’(x)sinf(x)
8.	Ktan f(x)	Kf’(x)sec2 f(x)
9.	Kcot f(x)	-Kf’(x)cosec 2f(x)
10.	Ksec f(x)	Kf’(x)sec f(x) tanf(x)
11.	Kcosec f(x)	Kf’(x)cosecf(x) cot f(x)
12.	Sin-1 f(x)
13.	Cos-1 f(x)
14.	Tan-1 f(x)
15.	Cosec-1 f(x)
16	Sec-1 f(x)
17.	Cot-1 f(x)

 

Table 1.8

Examples 1-9
	Function of Function	f(x)	Chain Rule
	y	u=	X
1.	e2x	2x	euX2	2e2x
2.	Inx3=loge x3	x3	(1/x3) X3x2	1/x
3.	a0.5x	0.5x
4.	Sin(x2+2x)	(x2+2x)
5.	Cos(Inx)	(Inx)
6.	Tan(logx)	(logx)	(secu)2X…..
7.	Cosec(3x3+2x2)	(3x3+2x2)	f(x) = 3x3+2x2
8.	Sec(5x2+4x )	(5x2+4x )	f(x) = 5x2+4x
9.	cot 50x	50x	f(x) = 50x

 

1.7 Function of a Function of a Function.

The relationship between the chain rule problems as discussed above and Function of a Function of a Function are as illustrated in the tables below.

The chain rule problems as stated in the table below have been raised to a power greater or less than 1.        

Table 1.9

Category C. A function of a function raised to a power greater than 1 or less than 1. Refer to Table 1.8
	Function of a Function	Exponent greater or less than 1	Function of a Function of a Function
			=   X X
2.	(Inx3)1= (loge x3)1	15	(Inx3)15 = (loge x3)15
4.	(Sin(x2+2x))1	19	(Sin(x2+2x))19
5.	(Cos(Inx))1	0.25	Cos(Inx))0.25
6.	(Tan(logx))1	-1/8	(Tan(logx))-1/8
7.	(Cosec(3x3+2x2))1	5	(Cosec(3x3+2x2))5
8.	(Sec(5x2+4x ))1	60	(Sec(5x2+4x ))60
9.	(cot 50x)1	20	(cot 50x)20

 

The second method of deriving a function of a function of a function is when the new f(x) is raised to a power greater than 1 or less than 1 as illustrated in the table below.

 

Table 1.10

Category D. A function of a function raised to a power greater than 1 or less than 1. See Table 1.8.
	Function of a Function	Exponent greater or less than 1	Function of a Function of a Function
			=   X
1.	In (x3)1= (loge(x3)1	50	In (x3)50 =loge(x3)50
2.	Sin(x2+2x)1	7	Sin(x2+2x)7
3.	Cos(Inx)1	0.5	Cos(Inx)0.5
4.	Tan(logx)1	1/3	Tan (logx)1/3
5.	Cosec(3x3+2x2)1	10	Cosec(3x3+2x2)10
6.	Sec(5x2+4x )1	9	Sec(5x2+4x )9
7.	cot (50x)1	4	cot (50x)4

 

The third method of deriving chain rule problems is a combination of the two modifications as illustrated in tables 1.9 and 1.10.

Table 1.11

Category E. A combination of the two previous tables. See Tables 1.9 and 1.10
	Function of a Function of  a Function	Function of a Function of  a Function	Function of a Function of a Function
			=   X X
1.	(Inx3)15 = (loge x3)15	In (x3)50 =loge(x3)50	(In (x3)50 )15 = (loge(x3)50)15
2.	(Sin(x2+2x))19	Sin(x2+2x)7	(Sin(x2+2x)7)19
3.	(Cos(Inx))0.25	Cos(Inx)0.5	(Cos(Inx)0.5)0.25
4.	(Tan(logx))-1/8	Tan (logx)1/3	(Tan (logx)1/3)-1/8
5.	(Cosec(3x3+2x2))5	Cosec(3x3+2x2)10	(Cosec(3x3+2x2)10)5
6.	(Sec(5x2+4x ))60	Sec(5x2+4x )9	(Sec(5x2+4x )9)60
7.	(cot 50x)20	cot (50x)4	(cot (50x)4)20

 

The next two categories are as illustrated in the following two tables.                                                                                                 

 

 

Table 1.12

Category F. The exponent has an x or x2 term.
	Function of a Function of  a Function	Function of a Function of  a Function	Function of a Function of a Function
			=   X X
1.	(Inx3)15x = (loge x3)15x	In (x3)50x =loge(x3)50x	(In (x3)50x )15x = (loge(x3)50x)15x
2.	(Sin(x2+2x))19x	Sin(x2+2x)7x	(Sin(x2+2x)7x)19x
3.	(Cos(Inx))0.25x	Cos(Inx)0.5x	(Cos(Inx)0.5x)0.25x
4.	(Tan(logx))(-1/8)x	Tan (logx)(1/3)x	(Tan (logx)x/3)-x/8
5.	(Cosec(3x3+2x2))5x	Cosec(3x3+2x2)10x	(Cosec(3x3+2x2)10x)5x
6.	(Sec(5x2+4x ))60x	Sec(5x2+4x )9	(Sec(5x2+4x )9x)60x
7.	(cot 50x)20x	cot (50x)4x	(cot (50x)4x)20x

 

When the exponent has a combination of x and y terms the differential of the problem may be solved implicitly.

 

 

 

 

Table 1.13

Category G. The exponent has a combination of x and y terms.
	Function of a Function of  a Function	Function of a Function of  a Function	Function of a Function of a Function
			=   X X
1.	(Inx3)15y = (loge x3)15y	In (x3)50x =loge(x3)50x	(In (x3)50x )15y = (loge(x3)50x)15y
2.	(Sin(x2+2x))19x	Sin(x2+2x)7y	(Sin(x2+2x)7y)19x
3.	(Cos(Inx))0.25y	Cos(Inx)0.5x	(Cos(Inx)0.5x)0.25y
4.	(Tan(logx))-x/8	Tan (logx)y/3	(Tan (logx)y/3)-x/8
5.	(Cosec(3x3+2x2))5x	Cosec(3x3+2x2)10y	(Cosec(3x3+2x2)10y)5x
6.	(Sec(5x2+4x ))60y	Sec(5x2+4x )9x	(Sec(5x2+4x )9x)60y
7.	(cot 50x)20x	cot (50x)4y	(cot (50x)4y)20y

 

The next category of function of a function are characterized by f(x) being replaced by a product of two or three functions as shown below.

 

 

 

Table 1.14

Category H. f(x) is a product of two or three functions.        f(x) = x is replaced with f(x) =uvw.
	Basic Function	Function of a Function of  a Function	Function of a Function of a Function
			The new function is raised to a power greater or lesser than 1
			Alternatively, the entire function is raised to a power greater or less than 1
			=   X X
1.	Inx	In(x(x2 +2)(x3+3))	In(x(x2 +2)(x3+3))3
2.	Sinx	Sin((x2+2x)(x3+3))	Sin((x2+2x)(x3+3))0.5
3.	Cosx	Cos((Inx)(logx))	Cos4 ((Inx)(logx))
4.	ex	e(cosx)(sinx)(cotx)	e2(cosx)(sinx)(cotx)
5.	2x	2(Inx)(logx)(cosx)	2100(Inx)(logx)(cosx)
6.	logx	log((cosx)(cotx)(cosx))	log((cosx)(cotx)(cosx))33
7.	Cotx	Cot((Inx)(logx)(ex))	Cot0.25((Inx)(logx)(ex))

 

 

 

 

Table 1.15

Category I. f(x)=x is substituted with an algebraic fraction consisting of two or more functions.
	Basic Function	Function of a Function of  a Function	Function of a Function of a Function
			The new function is raised to a power greater or lesser than 1
			Alternatively, the entire function is raised to a power greater or less than 1
			=   X X
1.	Inx	In[ ]	(In( ))50
2.	Sinx	Sin[ ]	Sin[ ]0.5
3.	Cosx	Cos[ ]	Cos4[ ]
4.	ex	e((cosx)(sinx))/(cotx)	e2((cosx)(sinx))/(cotx)
5.	2x	2((Inx)(logx))/(cosx))	2100((Inx)(logx))/(cosx)
6.	logx	log[ ]	log[ ]33
7.	Cotx	Cot[ ]	Cot0.25[ ]

 

1.8 Greatest and smallest functions                                            The differential of a function of a function based on the chain rule formulas as stated below shows that there is a smallest function and a greatest function.

Also, there may be an intermediate function depending on the problem given.

Given that y= In x³,                                                            =   X   u= x³is the smallest function while the greatest function is In x³.                                                                 

Table 1.16                                                               

Table **** Smallest Function and Greatest Function
		Smallest Function	Intermediate	Greatest Function
1.	(Inx3)15 = (loge x3)15	x3	Inx3	(Inx3)15 =(f(x))15
	Function	Cubic	Logarith-mic	Exponential
2.	(Sin(x2+2x))19	x2+2x	Sinx2+2x	(Sin(x2+2x))19
	Function	quadratic		Exponential
3.	(Cos(Inx))0.25	Inx	Cos(Inx)	(Cos(Inx))0.25
	Function	Logarith-mic		Exponential
4.	(Tan(logx))-1/8	logx	Tan u	Tan-1/8 (logx)
	Function	Logarith-mic		Exponential
5.	Cosec5(3x3+2x2)	3x3+2x2	Cosec u	Cosec5(3x3+2x2)
	Function	Cubic	Trigonom-etric	Exponential
6.	Sec60 (5x2+4x )	5x2+4x	Sec u	Sec60 (5x2+4x )
	Function	quadratic
7.	(cot 50x)20	50x	Cot u	(cot 50x)20
	Function	Straight Line	Trigonom-etric	Exponential

 

1.9 Chain Rule for Function of a Function of a Function.                                                                          As stated in the previous section, chain rule problems have smallest function, intermediate function and greatest function.                                                                               The table below illustrates the relationship between these functions and the chain rule as stated below.

 

 

 

Table 1.17                                                              

Table 1.17 Chain Rule    =   X   X
	differential	X X
1.	(Inx3)15 = (loge x3)15	=   X   X  =15u14 X  X 3x3 = 45u14m-1x3
	Substitution	y=u15	u= In m	m = x3
2.	(Sin(x2+2x))19	=   X   X  =19u18 XcosmX (2x+2) = 19u18(cosm)(2x+2)
	Substitution	y=u19	u= sinm	m= x2+2x
3.	(Cos(Inx))0.25	=   X   X  =0.25u-0.75 X-sinmX (1/x) = 0.25u-0.75(-sinm)(1/x)
	Substitution	y=u0.25	u=cosm	m= Inx
4.	(Tan(logx))-1/8	=   X   X  =(-1/8)u(-1/8)-1 Xsec2m X (1/(xIn10))
	Substitution	y=u-1/8	u= tan m	m=logx
5.	Cosec5(3x3+2x2)	=   X   X  =5u4 Xcosecmcotm X (9x2+4x) = 5u4(cosecmcotm)(9x2+4x)
	Substitution	y=u5	u=cosecm	u=3x3+2x2
6.	Sec60 (5x2+4x )	=   X   X  =60u59X(secmtanm)X (10x+4) = 60u59(secmtanm)( 10x+4)
	Substitution	y=u60	u=secm	u=5x2+4x
7.	(cot 50x)20	=   X   X  =20u19X(-cosec2 m)X (50) = 20u19(-cosec2 m)(50)
	Substitution	y=u20	u=cotm	u=50x

The differential of the problems in the table above is the product of the differential of greatest function, the derivative of the intermediate function and the differential of the smallest function.