Power of Maths Paper 1 OL

Techniques of Differentiation EXAMPLE 4

A rocket has a trajectory (path) described by the equation y = f (x) = 100x(6 − x).

Find the slope of the tangent to the curve at: (a) x = 1 (b) x = 3 (c) x = 5

y yf x=( )

Solution y = 100x(6 − x) = 600x − 100 x 2 dy

⇒

___ dx = 600 − 200x

(a) x = 1: (b) x = 3: (c) x = 5: x EXAMPLE If f (x) =

( 5x ___

Solution f (x) =

( 5x ___

2 − 3 , fi nd f ʹ(4). ) 2 − 3 )

f ʹ(x) = 25x ____

2

= 25 x 2 ____

Differentiate fi rst. f ʹ(x) = 25

2 − 15

Now substitute 4 for x: f ʹ(4) = 25 × 4

ACTIVITY ACTION CTION Differentiating algebraic expressions

OBJECTIVE To differentiate algebraic expressions and then to differentiate them again (second order differentiation)

Y 2

______ 2 − 15 = 35

Higher order differentiation A function may be differentiated many times. Consider y = f (x) = x 3 .

The fi rst derivative:

d x 2 is pronounced ‘Dee 2 y Dee x squared’. f ʹʹ is pronounced ‘f double dash x’. The slope m of a tangent to a curve is obtained by differentiating the equation of the curve once: Slope = m =

and so on. d 2 y

___

___ dx

dy This means that d 2y

___ dx2 is obtained by differentiating the slope m once:

d 2y

___ dx2 = dm

___ dx .

303

___ dx = f ʹ(x) = 3 x 2 is obtained by differentiating f (x) = x 3 .

dy The second derivative:

___ d x 2 = f ʹʹ(x) = 6x is obtained by differentiating f ʹ(x) = 3 x 2 ,

d 2 y

4 − 15x + 9 = 25 ___

___ 4 (2x) − 15(1 x 0 ) + 0

4 x 2 − 15 x 1 + 9 2 5

___ dx = 600 − 200 = 400

dy

___ dx = 600 − 600 = 0

dy

___ dx = 600 − 1000 = −400

dy

Can you explain these answers by reference to the diagram?

EXAMPLE 6 Find the derivative of y =

Solution Simplify fi rst.

y =

___ dx = 2x − 2(1 x 0 ) − 0 = 2x − 2 = 2(x − 1)

dy

____________ x + 3

( x 2 − 9) (x + 1) =

_________________ (x + 3)

(x − 3) (x + 3) (x + 1) y = (x − 3)(x + 1) = x 2 − 2x − 3

20

____________ x + 3

( x 2 − 9) (x + 1)

.

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