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Polynomials: Differentiation Rules
Use the Quotient Rule to differentiate r =
2 t2 – t + 6
---
2 t2 + 5 t – 5
dr
---
dt
 =
t4 + t3 + t2 + t +
---
t4 + t3 + t2 + t +

tickcross


Worked Solution:

The Quotient Rule:
y =
u(x)
---
v(x)
=>
dy
---
dx
 =
v
du
–--
dx
 – u
dv
---
dx
---
v2
Here, y = r, x = t, u = 2 t2 – t + 6 and v = 2 t2 + 5 t – 5
=>
du
---
dt
 = 4 t – 1 and
dv
---
dt
 = 4 t + 5
dr
---
dt
 =
( 2 t2 + 5 t – 5 )( 4 t – 1 ) – ( 2 t2 – t + 6 )( 4 t + 5 )
---
( 2 t2 + 5 t – 5 )2
=
12 t2 – 44 t – 25
---
4 t4 + 20 t3 + 5 t2 – 50 t + 25
or other equivalent forms.

© MEI Produced by Dr Ron Knott, revised 15 May 2007
tom . button [AT] mei . org . uk 		Test reference: PolyDiffRules.html?ref=56990,qu=quot