BC 1
Derivatives
Name:
The Chain Rule
This activity sheet is designed to help you to discover a general rule for differentiating the composition
of two or more functions. As a review, let’s consider some of the rules for derivatives we have seen so
far that involve variable expressions other than just x. Fill in each of the following (assume k and n are
constants).
y = (x + k)n

y =
y = (kx)n

y =
y = e xk

y =
y = e kx 

y =
y = ln(x + k)

y =
y = ln(kx)

y =
y = sin(x + k) 
y =
y = sin(kx)

y =
Now let’s step things up a bit. Fill in each blank with what you believe will be the derivative of the
given function. Then in the space provided, write the derivative provided by the TI-89. Hopefully they
will match, eventually.
Your derivative
(1.1)
y = (3x + 4)10
(1.2)
y
TI-89 derivative
y =
y =
 5 x  7 4
y =
y =
(1.3)
y = (x2 – 11x + 1)5
y =
y =
(1.4)
y  x3  1
y =
y =
(2.1)
y = ln(3 – 4x)
(2.2)
y  e4 x
(2.3)
y  sin
(2.4)
y  cos x3  4 x 2  1
IMSA BC 1
2
6 2 x
 x
4

3


y =

y =

y =

y =
y =
y =
y =
y =
Chain Rule.1
Rev. F07
BC 1
Derivatives
Name:
The Chain Rule
Your derivative
TI-89 derivative
(3.1)
y = (ln (x))5
y =
y =
(3.2)
y  3 cos  x 
y =
y =
(3.3)
y
e 
y =
y =
(3.4)
y = 5 sin3(2x)
y =
y =
(4.1)
y = etan x
y =
y =
(4.2)
y =ln(cos x)
y =
y =
(4.3)
y  sec e x
 
y =
y =
(4.4)
y = sin4(ln(x))
y =
y =
10
x 4
Given what you have seen so far, write a description of how you could take the derivative of the
composite of two functions. Then complete the statement at the bottom of the page.
Chain Rule: If k(x) = ƒ(g(x)), then k(x) =
IMSA BC 1
Chain Rule.2
Rev. F07
Now it’s your turn! Use the chain rule to differentiate each of the following functions.
Do not simplify.
(1)
y = (2x + 7)14

y =
(2)
y = 5(x3 + 1)–2

y =
(3)
y = 2(3x5 – 6x2 + 18)23

y =

y =
(4)
2
 1

y   x 4  3 x 3  1


5
(5)
y= 2 x

y =
(6)
y = ln(4x2 + 3x – 1)

y =
(7)
y = (tan x)–4

y =
(8)
y = 4 sec5(x)

y =
(9)
y = cos(5x2 – 3x + 2)

y =
IMSA BC 1
Chain Rule.3
Rev. F07
( x 2  1)3
4x  3
(10)
y=
(11)
y = (5x – 4)3 · (x4 + 1)2

y =

y =

y =
y =
(12)
y = cot2(x) · tan 4x
y =
(13)
y=
sin 3  x 
cos 4  x 
y =
(14)
(15)
y = (ln(6x + 2))3
y= e
IMSA BC 1
 
cos x 2
Chain Rule continued.2
Rev. F07