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+.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
+