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Revision Notes for Core 3

Functions

Domain is the set of inputs for a function. Can be all real numbers or only a particular

few.

Range is the set of output numbers obtained from the domain. Use completing the

square to find the lowest and highest values for the range if a quadratic.

Composite functions

gf(x) = g(f(x)) means substitute in x into function f then substitute in that result into

function g. It is a function of a function.

Inverse function means just doing the function in reverse

-1

f

(x) when f(x) = (2x+3)/5 becomes (5x-3)/2. Think of as a flow diagram followed in

reverse. The range and domain swap around when doing the inverse and it is a

reflection in the line y=x if they are one-to-one functions (one value for the range for one

value from the domain)

n

Differentiating (ax+b)

Use the chain rule

dy/dx = dy/du x du/dx where u=ax+b Use whenever you have a function of a function.

Diff one function then diff the other and multiply the results together. Use for rates of

change as well to find dy/dt for instance.

Product Rule

Quotient Rule

dy

du

dv

u

dy

du

dv

If y=uv then

= v

+ u

If y=

then

= v

-u

dx

dx

dx

v

dx

dx

dx

dy

2

= ( diff leave + leave diff )

v

dx

Make sure u is the numerator and v the

denominator.

d

1

2

2

e.g

x

lnx = 2x lnx + x

x

dx

n

Integrating (ax+b)

1

1

∫

+ ) b

n

n+1

(

ax

dx =

x

(ax+b)

+ c

This is because if you differentiate

+

n

1

a

n+1

n

(ax+b)

you get: a x (n+1) (ax+b)

so

1

1

the

x

counteracts this.

+

n

1

a

x

Differentiating and integrating e

and ln x

d

x

x

e

= e

as by definition the gradient is the same as the function and therefore

dx

d

1

x

x

3x

3x

3x

3x

∫ e

and ∫ e

dx = e

+ c

e

use chain rule so = 3e

dx =

e

+ c for the same

dx

3

reason as the reverse of the chain rule.

Log Laws

a

x

ln a + ln b = ln ab

ln a – ln b = ln

ln 1 = 0

ln a

= x ln a

b

x

Remember that if y = e

then if I take the natural log (ln) of both sides, then x = ln y

x

x

(e

and ln x are the inverse of each other).

LEARN: y = e

so x = ln y

d

1

1

and ∫

ln x =

dx = ln x + c

dx

x

x

1

1

1

1

If you can take out a coefficient then do ∫

∫

dx =

dx =

ln x + c

x

3

x

3

3

1

1

1

If of the form ∫

dx then =

ln (ax+b) as

counteracts the differentiation of ax+b.

ax + + + +

b

a

a

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