.2C

.NH
Differentiating standard equations
.EQ L
dy over dx = n x sup n-1 
.EN
.NH 2
Example
.LP
Define the equation
.EQ L
y = 5 x sup 2 + 19 x + 8
.EN
Use the formula
.EQ L
dy over dx = 10 x + 19
.EN
Its very simple
.NH
Differentiating trig equations
.LP
From the chain rule, one can find the following:
.EQ L
sin(kx) -> k cos(kx)
.EN
.EQ L
cos(kx) -> -k sin(kx)
.EN
.EQ L
tan(kx) -> k sec sup 2 (kx)
.EN
.EQ L
sec(kx) -> k sec(kx) tan(kx)
.EN
.EQ L
cot(kx) -> -k cosec sup 2 (kx)
.EN
.EQ L
cosec(kx) -> -k cosec(kx) cot(kx)
.EN
.NH
Chain rule
.EQ L
dy over dx = dy over dt times dt over dx
.EN
.NH 2
Example
.LP
Define the function
.EQ L
y =sin sup 2 (9x)
.EN
Re-write y in terms of t
.EQ L
Y =sin sup 2 (t)
.EN
Define t
.EQ L
t = 9x
.EN
Differentiate y with respect to t
.EQ L
dy over dt = 2cos(t)
.EN
Differentiate y with respect to x
.EQ L
dt over dx = 9
.EN
Times the two together
.EQ L
dy over dx = 2cos(t) times 9
.EN
Substute the original t back in
.EQ L
dy over dx = 18cos(9x)
.EN

.NH
Product rule

.EQ L
"when " y = u v
.EN
.EQ L
dy over dx = ( v prime times u ) + ( v times u prime )
.EN
.NH 2
Example

.LP
Define the equation
.EQ L
y = sin(x) cos(x)
.EN
Define u and v
.EQ L
u = sin(x)
.EN
.EQ L
v = cos(x)
.EN
Differentiate indiviually
.EQ L
u prime = cos(x)
.EN
.EQ L
v prime = -sin(x)
.EN
Put into the formula
.EQ L
( v prime times u ) + ( v times u prime ) = -sin(x)sin(x) + cos(x)cos(x)
.EN
Simplify
.EQ L
dy over dx = -sin sup 2 (x) +cos sup 2 (x)
.EN
.EQ L
dy over dx = cos(2x)
.EN