1. DIFFERENTIATION
TECHNIQUES

2. with the following standard results:
To differentiate various expressions it is necessary to be very familiar
with the following standard results:
Trigonometric:
Exponential: y dy dx y dy dx
sin kx
k cos kx
e kx
ke kx
cos kx
–k sin kx
Logarithmic:
tan kx
k sec2 kx y dy dx
sec x
sec x tan x
cosec x
– cosec x cot x
f (x)
f(x)
ln f(x)
cot x
– cosec2x

3. It is also necessary to know the following:dy dx =
Products:
If y = uv, then
uv + vu
If y = u v
, then dy dx =
vu – uv v2
Quotients:
“Bracket to a power”
If y = [ f(x) ] n
, then dy dx =
n [ f(x) ]
n-1
f (x)

4. dy dx Example 1: Find given that y = x2 sin 4x dy dx =
u v dy dx =
This is a product, so x2
( 4 cos 4x ) +
sin 4x
( 2x ) u v v u
= 2x ( 2x cos 4x + sin 4x )
Example 2: Find dy dx
given that y = x
e2x u v v u u v dy dx =
e2x
(1) – x
(2e2x)
This is a quotient, so
(e2x)2 v2 dy dx =
1 – 2x
e2x
e2x :
Divide throughout by

5. Example 3: Find dy dx
given that y = 3 ( 2x + 1 )5.
This is a “bracket to a power”.
f (x) dy dx =
15( 2x + 1 )4
× 2
= 30 ( 2x + 1 )4
Example 4: Find dy dx
given that
y = ln ( 1 + 2x2)
f (x) dy dx = 4x
f(x)
1 + 2x2

6. If x is given as a function of y, it can sometimes be necessary to use
the result: dy
d x
d y 1 =
Example 5: Given that x = tan y, find dy
d x
in terms of x.
x = tan y
d x
d y =
sec2 y dy
d x = 1
sec2 y = 1
1 + tan2 y = 1
1 + x2

7. It is vital to know the following results:
Summary of key points:
It is vital to know the following results:
Trigonometric:
Exponential: y dy dx y dy dx
sin kx
k cos kx
e kx
ke kx
cos kx
–k sin kx
Logarithmic:
tan kx
k sec2 kx y dy dx
sec x
sec x tan x
cosec x
– cosec x cot x
f (x)
f(x)
ln f(x)
cot x
– cosec2x

8. ……and the following: dy dx = Products: If y = uv, then uv + vu If y =
vu – uv v2
Quotients:
“Bracket to a power”
If y = [ f(x) ] n
, then dy dx =
n [ f(x) ]
n-1
f (x)
This PowerPoint produced by R.Collins ; Updated Apr.2009

9. This is an example of “implicit differentiation”.dy dx
Example 6: Find
given that x2 – y3 = 5x + y
This is an example of “implicit differentiation”.
The method is to differentiate each term with respect to x.
However the chain rule is needed for the terms involving y. d dx
f(y) = dy
f(y)
i.e. use the result:
In practice, this means differentiating with respect to y and
multiplying by dy dx .
– 3y2 dy dx dy dx +
So: 2x
= 5 dy dx =
2x – 5
1 + 3y2
Re-arranging this gives: