Cambridge Additional Mathematics

Introduction to differential calculus (Chapter 13) 345

The rules you found in theDiscoveryare much more general than the cases you just considered.

For example, if f(x)=xn then f^0 (x)=nxn¡^1 is true not just for all n 2 Z+, but actually for all

n 2 R.

We can summarise the following rules:

f(x) f^0 (x) Name of rule

c(a constant) 0 differentiating a constant

xn nxn¡^1 differentiatingxn

cu(x) cu^0 (x) constant times a function

u(x)+v(x) u^0 (x)+v^0 (x) addition rule

The last two rules can be proved using the first principles definition of f^0 (x).

² If f(x)=cu(x) wherecis a constant, then f^0 (x)=cu^0 (x).

Proof: f^0 (x) = lim

h! 0

f(x+h)¡f(x)

h

= lim

h! 0

cu(x+h)¡cu(x)

h

= lim

h! 0

c

h

u(x+h)¡u(x)

h

i

=clim

h! 0

u(x+h)¡u(x)

h

=cu^0 (x)

² If f(x)=u(x)+v(x) then f^0 (x)=u^0 (x)+v^0 (x)

Proof: f^0 (x) = lim

h! 0

f(x+h)¡f(x)

h

= lim

h! 0

³

u(x+h)+v(x+h)¡[u(x)+v(x)]

h

́

= lim

h! 0

³

u(x+h)¡u(x)+v(x+h)¡v(x)

h

́

= lim

h! 0

u(x+h)¡u(x)

h

+ lim

h! 0

v(x+h)¡v(x)

h

=u^0 (x)+v^0 (x)

Using the rules we have now developed we can differentiate sums of powers ofx.

For example, if f(x)=3x^4 +2x^3 ¡ 5 x^2 +7x+6 then

f^0 (x) = 3(4x^3 ) + 2(3x^2 )¡5(2x) + 7(1) + 0

=12x^3 +6x^2 ¡ 10 x+7

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_13\345CamAdd_13.cdr Tuesday, 7 January 2014 9:52:53 AM BRIAN