Examples of calculation of derivative

See Derived for more total information.

The derived is a mathematical function, more precisely a function of functions because it takes as argument of entry a function and returns another function, generally different.

Examples starting from the definition of the derived number

Constant function

That is to say C a reality.

Let us consider the constant function F of value C :

\ forall X \ in \ mathbb {R}, \ forall H \ in \ mathbb {R^*}, \ frac {F (x+h) - F (X)}{H} = \ frac {DC} {H} = 0

thus
\ forall X \ in \ mathbb {R}, f' (X) = \ lim_ {H \ rightarrow 0} \ frac {F (x+h) - F (X)}{H} =0.

Thus the derivative of a constant function is the null function.

Function square

Let us consider the function F definite on \ mathbb {R} by
\ forall X \ in \ mathbb {R}, F (X) =x^2

\ forall X \ in \ mathbb {R}, \ forall H \ in \ mathbb {R} ^*, \ frac {F (x+h) - F (X)}{H} = \ frac {(x+h) ^2-x^2} {H}

= \ frac {(x+h-x) (x+h+x)}{H} = \ frac {H (2x+h)}{H} =2x+h

thus
f' (X) = \ lim_ {H \ rightarrow 0} (2x + H) =2x
the derivative of F is thus the function F ' defined by
\ forall X \ in \ mathbb {R}, f' (X) =2x.

Function root

Let us consider the function F =√x

\ forall X \ in \ mathbb {R} _+^*, \ forall H \ in \ mathbb {R} ^*, h>-x, \ quad \ frac {F (x+h) - F (X)}{H} = \ frac {\ sqrt {x+h} - \ sqrt {X}} {H}

= \ frac {(\ sqrt {x+h} - \ sqrt {X}) (\ sqrt {x+h} + \ sqrt {X})}{H (\ sqrt {x+h} + \ sqrt {X})}

= \ frac {x+h - X} {H (\ sqrt {x+h} + \ sqrt {X})}= \ frac {1} {\ sqrt {x+h} + \ sqrt {X}}

thus

\ forall X \ in \ mathbb {R} _+^*, f' (X) = \ lim_ {H \ rightarrow 0} \ frac {1} {\ sqrt {x+h} + \ sqrt {X}} = \ frac {1} {2 \ sqrt {X}}

In addition,

\ forall H \ in \ mathbb {R} _+^*, \ frac {F (H) - F (0)}{H} = \ frac {\ sqrt {H}} {H} = \ frac {1} {\ sqrt {H}}

\ lim_ {H \ rightarrow 0} \ frac {F (H) - F (0)}{H} =+ \ infty

thus F is not derivable into 0 and the representative curve admits into 0 a half vertical tangent.

Examples starting from the formulas of derivative

Here a series of examples of derivative calculated starting from the formulas established by the method with the limit.

Second degree

Let us consider the following functions and then derive them thereafter:

1. y=x^2+5x-3 \,

2. y=3x^2-9x+ \ frac {2} {3} \,

3. y=-4x^2+ \ frac {4} {7} x-1 \,

Derived: 1. y=x^2+5x-3 \,

y'= (x^2) “+ (5x)” - (3) '\,

y'=2x+5+0 \,

y'=2x+5 \,

2. y=3x^2-9x+ \ frac {2} {3} \,

y'= (3x^2) '- (9x) “+ (\ frac {2} {3})” \,

y'=6x-9+0 \,

y'=6x-9 \,

3. y=-4x^2+ \ frac {4} {7} x-1 \,

y'= (- 4x^2) “+ (\ frac {4} {7} X)” - (1) '\,

y'=-8x+ \ frac {4} {7} - 0 \,

y'=-8x+ \ frac {4} {7} \,

Third degree

Let us consider the following functions and derive them thereafter:

1. y=2x^3+6x^2-4x+ \ frac {9} {\ pi} \,

2. y=-x^3-5x^2+ \ frac {2} {3} x-1 \,

3. y= \ frac {5} {17} x^3+x^2-2x+e \,

Derived:

1. y=2x^3+6x^2-4x+ \ frac {9} {\ pi} \,

y'= (2x^3) “+ (6x^2)” - (4x) “+ \ left (\ frac {9} {\ pi} \ right)” \,

y'=6x^2+12x-4+0 \,

y'=2 (3x^2+6x-2) \,

2. y=-x^3-5x^2+ \ frac {2} {3} x-1 \,

y'=- (x^3) “- (5x^2) “+ \ left (\ frac {2} {3} X \ right)” - (1)” \,

y'=-3x^2-10x+ \ frac {2} {3} - 0 \,

y'=-3x^2-10x+ \ frac {2} {3} \,

3. y= \ frac {5} {17} x^3+x^2-2x+e \,

y'= \ left (\ frac {5} {17} x^3 \ right) “+ (x^2)” - (2x) “+ (E)” \,

y'= \ frac {5 \ times 3x^2} {17} +2x-2+0 \,

y'= \ frac {15x^2} {17} +2x-2 \,

Unspecified degree

Are the general forms:
1) \ qquad y=f (X) =ax^b \ qquad (has \ not=0, B \ in \ mathbb {N^*})

Nth derivative:

  • \ forall N \ in \ mathbb {NR} \ qquad n

  • \ forall N \ in \ mathbb {NR} \ qquad n=b: \ qquad y^ {(N)}=a B! \,

  • \ forall N \ in \ mathbb {NR} \ qquad n>b: \ qquad y^ {(N)}=0 \,

2) \ qquad y=f (X) =ax^ {- B} \ qquad (has \ not=0, B \ in \ mathbb {N^*})

Nth derivative:

  • \ forall N \ in \ mathbb {NR}: \ qquad y^ {(N)}=a (- 1) ^n \ frac {(b+n-1)!}{(b-1)! x^{n+b}}

Random links:Perwez | Umberto Brunelleschi | Auzat | Beryl Markham | Bitty Schram | Dahus | Dacula,_la_Géorgie