Tangent to a Curve & Derivative of a Function (8.1 – 8.9)

 

8.1.            Informal discussion of continuity.

8.2.            The notion of the limit of a function and the definition of continuity in terms of this notion. Continuity of f + g, f – g, fg in terms of continuity of f & g.

8.3.            Gradient of a secant to the curve y = f(x).

8.4.            Tangent as the limiting position of a secant. The gradient of the tangent. Equations of tangent and normal at a given point of the curve y = f(x).

8.5.            Formal definition of the gradient of y = f(x) at the point where x = c.           Notations f ’ (c), at x = c.

8.6.            The gradient or derivative as a function.

Notations f ’(x), , (f(x)), y’.

8.7              Differentiation of xn for positive integral n. The tangent to y = xn.

8.8              Differentiation of x½ and x-1 from first principles. For the two functions u and v, differentiation of Cu (C constant), u + v, u – v, uv. The composite function rule. Differentiation of u/v.

8.9              Differentiation of: general polynomial, xn  for n rational, and functions of the form {f(x)}n or f(x)/g(x), where f(x), g(x) are polynomials.”  (syllabus)

 

 

 

 

 

 

 

 

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LIMITS

 

Q.1.     Find the values of the following limits:

(i)           (ii)            (iii)            (iv)

(v)             (vi)                 (vii)

(viii)        (ix)              (x)

 

Q.2.     Find the values of the following limits:

(i)               (ii)             (iii)

(iv)             (v)               (vi)

(vii)       (viii)        (ix)

(x)

 

Q.3.     Find the values of the following limits:

(i)                  (ii)                   (iii)

(iv)            (v)                          (vi)

(vii) (viii)

(ix)            (x)

Q.4.     Explain why  does not exist.

Q.5.     Does  exist? If so, what is its value?

Q.6.     Does exist? If so, what is its value?

 

Answers:

Q.1.     (i) 4,  (ii) 2,  (iii) 0,   (iv) 0,  (v) 0,  (vi) 36,  (vii) 32,  (viii) -4,  (ix) 0,  (x) -54

Q.2.     (i) -1,  (ii) 4,  (iii) -1,  (iv) 6,  (v) 12,  (vi) 1,  (vii) 3,  (viii) ,  (ix) 11,  (x) -8

Q.3.     (i) 1,  (ii) ,  (iii),  (iv) ,  (v) 0,  (vi) 0,  (vii) 0,  (viii),  (ix) ,  (x) 1

Q.4.     As x0,  .      is not a number so  does not exist.

Q.5.     Does not exist.              Q.6. Does not exist.

CALCULUS- FIRST PRINCIPLES

Suppose we want to determine the gradient between two points on the curve y = x2.

Let the first point be (x, y)

Consider another nearby point and let dx be the increase in the x value and dy be the increase in the y value.

The co-ordinates of the new point are (x + dx, y + dy)

The average gradient between the points (x, y) and (x + dx, y + dy) is

 

           

            All points on the parabola satisfy the equation of the parabola.

            So y = x2

            And y + dy = (x + dx)2

            i.e. y + dy = x2 + 2xdx + dx2

                Subtract y from the LHS and x2 from the RHS (they are equal)

            dy = 2xdx + dx2

            dy = dx(2x + dx)

            = 2x + dx

           

This means that the gradient between two points that are very, very close together on the parabola y = x2 is equal to 2x.

 

It can similarly be shown that the gradient between two very close points on a curve is equal to:

                                    =

Exercise: Find  from first principles for the following curves:

(i) y = 2x2         (ii) y = 3x2        (iii) y = 2x2 + 3 (iv) y = 2x2 + x

(v) y = 2x2 – x              (vi) y = - 2x2     (vii) y = 2x3      (viii) y = 4x2 -3x

(ix) y = 4x2 – 3x + 2     (x) y = 4x2 – 3x – 5

Answers:

(i)      4x,  (ii) 6x,  (iii) 4x,  (iv) 4x + 1,  (v) 4x – 1,  (vi) -4x,  (vii) 6x,  (viii) 8x – 3, 

(ix) 8x – 3,  (x) 8x - 3  

 

DIFFERENTIATION

The  is known as the differential and is written .

The process of finding the differential is known as differentiation.

You can show from first principles that the differential of cxn is ncxn-1

If y = cxn  then = ncxn-1
 

 

e.g. y = 5x2       = 10x

To differentiate a constant we differentiate cx0 and get 0.

To differentiate the addition of different powers of x, each term is differentiated separately.

If y = cxn + kxm + g  then = ncxn-1 + mkxm-1

 

 
 

 

                        e.g. y = 7x3 – 4x + 8      = 21x2 – 4

 

Exercise 1. Differentiate the following expressions with respect to x.

(i) 5x2               (ii) 7x3              (iii) 12x5           (iv) 8x2 + 2x + 3           (v) 2x3 + 4x – 2

(vi) 2x4 – 7x + 4           (vii) 2x4 -7x – 12          (viii) 4x3 – 8x2 – 3x + 2

(ix) 8x12 – 7x6               (x) x5 + 4x3 + 2x - 3

 

Exercise 2. Differentiate the following expressions with respect to x.

(i) x -2               (ii)              (iii)             (iv)            (v)              (vi)

(vii)           (viii) (ix)           (x)

 

Answers:

Ex.1.    (i) 10x,  (ii) 21x2,  (iii) 60x4,  (iv) 16x + 2,  (v) 6x2 + 4,  (vi) 8x3 – 7, 

(vii) 8x3 -7,  (viii) 12x2 – 16x -3,  (ix) 96x11 – 42x5,  (x) 5x4 + 12x2 + 2

Ex.2.    (i) -2x -3 or            (ii) -3x -4 or                        (iii)  or        

            (iv)  or           (v) -10x -6 or        (vi)  or

(vii) or       (viii)       (ix)  or

            (x)  or  

 

 

 

THE CHAIN RULE

 

Consider the equation: y = (x2 + 3)3

One way to differentiate is to expand the RHS and then differentiate.

(x2 + 3)3 = x6 + 9x4 + 27x2 + 27

 = 6x5 + 36x3 + 54x

An alternative method is to use the chain rule:  = .

In the above example we would first differentiate (x2 + 3)3 with respect to (x2 + 3) and

then differentiate (x2 + 3) with respect to x.

 = .

            = 3(x2 + 3)2. 2x

            = 6x(x2 + 3)2

Normally we would leave the answer in this form but in this case we will expand it to show that the answer is the same as above.

=  6x(x2 + 3)2

            = 6x(x4 + 6x2 + 9)

            = 6x5 + 36x3 + 54x       (the same as above)

The Chain Rule: = .
 

 

Q.1. Differentiate the following expressions first by expanding and differentiating and then by the chain rule. Show that the answers are the same by each method.

         (i) (x2 + 4)2          (ii) (x3 +2)2                   (iii) (5x + 4)3                 (iv) (2x2 - 3)2   

(v) (5x5 + 4)2       (vi) (x2 - 3x)3                (vii) (10x6 + 2x)2           (viii) (6x2 + 5)3

(ix) (8x4 + 3x)2    (x) (12x3 + 4x2)2

Q.2.  Use the chain rule to differentiate the following expressions:

         (i) (x2 - 4)2           (ii) (x3 - 6)2       (iii) (6x2 + 2x)5   (iv) (x2 – 6)27  

(v) (12x8 + 4x3)8              (vi) (8x16 – 7x13)11         (vii)     (viii)

(ix)         (x)

Answers:

Q.1.  (i) 4x(x2 + 4),   4x3 + 16x                        (ii) 6x2(x3 + 2),   6x5 + 12x2

         (iii) 15(5x + 4)2,   375x2 + 600x + 240    (iv) 8x(2x2-3),   16x3 – 24x

         (v) 50x4(5x5 + 4),   250x9 + 200x4          

(vi) 3(2x-3)(x2 – 3x)2,  6x5 – 45x4 + 108x3 – 81x2

         (vii) 2(10x6 + 2x)(60x5 + 2),  1200x11 + 280x6 + 8x

         (viii) 36x(6x2 + 5)2,   1296x5 + 2160x3 + 900x

         (ix) 2(8x4 + 3x)(32x3 + 3),  512x7 + 240x4 + 18x

         (x) 2(12x3 + 4x2)(36x2 + 8x),   864x5 + 480x4 + 64x3

Q.2.  (i) 4x(x2 - 4)        (ii) 6x2(x3 -6)    (iii) 5(12x + 2)(6x2 + 2x)4         (iv) 54x(x2 – 6)26

   (v) 8(96x7 + 12x2)(12x8 + 4x3)7            (vi) 11(128x15 – 91x12)(8x16 – 7x13)10                (vii)    (viii)           (ix)          (x)

THE PRODUCT RULE

 

The differential of the product of two functions of x, u and v is given by:

 

 

Example: Consider the function y = x2(x3 + 3)

 

One way of finding the differential would be to multiply the function and then to differentiate.

y = x2(x3 + 3) = x5 + 3x2

= 5x4 + 6x

 

Another way of finding the differential is to use the product rule

 = x2  + (x3 +3)         

= x2(3x2) + (x3 +3)2x

= 3x4 + 2x4 + 6x

= 5x4 + 6x

 

Exercise 1: Differentiate the following functions of x by (a) multiplying and then differentiating, (b) the chain rule and (c) the product rule. Show that the answers are the same by each method.

(i) y = (x + 2)2 (ii) y = (x2 + 4)2            (iii) y = (2x + 3)2           (iv) y = (x2 + 2x)2

(v) y = (x3 + 5x2)2

 

Exercise 2: Use the product rule to differentiate the following products:

(i) y = x3 (2x + 8)         (ii) y = 2x (x3 + 7)         (iii) y = x5 ( 2x3 – 8)      (iv) y = 5x4(x8 -7)

(v) y = (x - 3)(x2 + 3x + 9)       (vi) y = (2x – 4)(3x + 7)           (vii) y = (x2 – 18)(x5 + 7)

(viii) y = (x2 + 4x)(x7 – 3x6)       (ix) y = (8x – 4)( 3x – 2x2)        (x) y = (x3 + 27)(x3 – 27)

 

Exercise 3:

(i)                  If f(x) = (2x + 4)( x – 3), find f ’(2)

(ii)                If f(x) = (5x - 4)( x2 + 7), find f ’(5)

(iii)               If f(x) = (x2 + 8)( 2x3 – 5), find f ’(3)

(iv)              If f(x) = (2x3- 16)( 9x8 – 8x2), find f ’(2)

(v)                If f(x) = (6x3 + 4x)( 2x2– 6x), find f ’(1)

 

Answers:

1. (i) 2x + 4      (ii) 4x3 + 16x    (iii) 8x + 12      (iv) 4x3 + 12x2 + 8x      (v) 6x5 + 50x4 + 100x3

2. (i) 8x3 – 24x2,  (ii) 8x3 + 14,  (iii) 16x7 – 40x4,  (iv) 60x11 – 140x3,  (v) 3x2, 

    (vi) 12x + 2   (vii) 7x6 – 90x4 + 14x,  (viii) 9x8 + 8x7 – 84x6,  (ix) -48x2 + 64x – 12

     (x) 6x5

3.  (i) 6,  (ii) 370,  (iii) 1212,  (iv) 54528,  (v) -108

 

 

THE QUOTIENT RULE

 

The differential of the quotient of two functions of x, u and v is given by:

 

 

Example: Find the derivative of the function: y =

 

 =  

 

=  =  =

 

Exercise: Use the quotient rule to find the derivative of the following functions:

(i) y =                (ii) y =               (iii) y = (iv) y =

(v) y =            (vi) y =         (vii) y =       (viii) y =

(ix) y =           (x) y =          (xi) y =           (xii) y =

(xiii) y =        (xiv) y =        (xv) y =        (xvi) y =

Answers:

(i)        (ii)    (iii)       (iv) (v)   (vi)

(vii)            (viii)    (ix)         (x)       (xi) 1   

(xii)             (xiii)    (xiv) 2x + 3      (xv)

(xvi)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(iv) y = 2 – 4x – 2x2 (v) y = 3x3 + 4x2 – 6x - 18           

(vi) y = 8 – 3x + 4x2 – 2x3