Transcript “Teach A Level Maths” Vol. 2: A2 Core Modules 9a: Differentiating Harder Products © Christine Crisp.

“Teach A Level Maths”
Vol. 2: A2 Core Modules
9a: Differentiating Harder
Products
© Christine Crisp
Differentiating Harder Products
Module C3
OCR
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Differentiating Harder Products
We often find that one of the factors of a product
is a function of a function.
e.g. 1 Find dy if y  x 2 e 3 x
Solution:
dx
Let u  x 2 and v  e 3 x
du
 2x
dx
dy
dv
du
u
v
dx
dx
dx
dv
 3e 3 x
dx
dy
 3x 2 e 3x  2xe 3x
dx
Differentiating Harder Products
e.g. 2 Find dy if
y  e x (1  x 2 ) 3
dx
We can set out the solution as follows:
Let
x and
ue
du
ex
dx
v  (1  x 2 ) 3
w  1  x2  v  w3
dv
dv
 6 x (1  x 2 ) 2 dw  2 x
 3w 2
dx
dw
dx
 3(1  x )
dy
dv
du
u
v
dx
dx
dx

2 2
dv

 6 x (1  x 2 ) 2
dx
dy
x
2 2
x
2 3
  6xe (1  x )   e (1  x )
dx
x
2 3
x
2 2
 e ( 1  x )  6x e ( 1  x )
 e (1  x ) ( 1  x  6 x )
x
2 2
2
Differentiating Harder Products
Differentiating Harder Products
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Differentiating Harder Products
We often find that one of the factors of a product
is a function of a function.
Differentiating Harder Products
Find dy if
e.g.
y  e x (1  x 2 ) 3
dx
We can set out the solution as follows:
Let
u  e x and
du
ex
dx
v  (1  x )
2 3
w  1  x2  v  w3
dv
dv
2
 6 x (1  x 2 ) 2 dw  2 x

3w
dx
dx
dw
 3(1  x )
dv

 6 x (1  x 2 ) 2
dx
dy
x
2 3
x
2 2
 e (1  x )   xe (1  x )
dx
 e x ( 1  x 2 ) 3  x e x( 1  x 2 ) 2
x
2 2
 e (1  x ) ( 1  x 2  x )
2 2


