Section 3.4 – The Chain Rule.
Here are some examples of using the chain rule to find derivatives.
Page 205 # 10: f(x) =
1
(1 + secx)2
Rewrite this as: f(x) = (1 + secx) -2 so that we don’t have to use the quotient rule as well as
the chain rule.
f(x) = ()-2 where ( )-2 is the outside function and  is the inside function,
where  = 1 + secx.
So f’(x) = -2()-3 ()’
we have
(That’s the chain rule.) Since  = 1 + secx, ()’ = secx tanx and
f ’(x) = -2(1 + secx) -3 (secx tanx)
Some people may prefer to write the steps a bit differently. I will show that on the board.
#20: F(t) = (3t – 1)4 (2t + 1)-3
 v 6
#30: F(v) =  3

v +1
#40: y = sin(sin(sinx))
#50: y = e
ex
Now it is time for you to try some.
t
#12: f(t) = sin(e ) + e
sint
2
#22(sort of):
f(w) =
2
#32: y = sec (mθ)
w +1
2
w +4
#42: y =
x+
x+
x
#52: Find an equation of the tangent line to the curve y =
1+x
2
at the point (2,3)
#60: Find the x-coordinates of all points on the curve y = sin2x – 2sinx at which the tangent
line is horizontal.
If the tangent line is horizontal, then the slope of the tangent line is zero, so we are looking
for all the points on the curve where the derivative (which is the function that will give us the
slope) is zero.
y = sin2x – 2sinx →
dy
dy
= cos2x (2x)’ – 2cosx →
= 2cos2x – 2cosx
dx
dx
dy
= 0 → 2cos2x – 2cosx = 0 → cos2x – cosx = 0
dx
You may remember how to solve this from your trigonometry or precalculus class, but there is
a good chance that you don’t remember correctly. We have to write both trig functions in
2
terms of the same argument, so we will use the double angle formula cos2x = 2cos x – 1.
2
2cos x – 1 – cosx = 0. This is an equation of quadratic form. Using the most powerful tool of
2
mathematics, we will let cosx = w and we have: 2w – w – 1 = 0. [You do not have to do
this substitution if you do not wish to do so.]
This equation factors nicely: (2w + 1)(w – 1) = 0, so we have w = -1/2 or w = 1.
w = 1 → cosx = 1 → x = 2k where k is an integer.
-1
-1
w = 2 → cosx = 2
→ x= ±
2
2
+2k → x =
(3k ± 1) where k is an integer.
3
3
The x-coordinates are numbers of the form x = 2k or x =
2
(3k ± 1) where k is an integer.
3
#76: For what values of r does the function y = e
rx
satisfy the differential equation
y’’ – 4y’ + y = 0?
y= e
rx
,
so y’ =
and y’’ =
Putting these into the equation y’’ – 4y’ + y = 0 gives us:
2 rx
r e
- 4re
rx
+e
rx
2
r – 4r + 1 = 0.
So
r=2±
3
=0
rx
2
or e (r – 4r + 1) = 0
rx
Since e ≠ 0 we need to solve
#94: Suppose y = f(x) is a curve that always lies above the x-axis and never has a horizontal
tangent, where f is differentiable everywhere. For what value of y is the rate of change of y
with respect to x eighty times the rate of change of y with respect to x?
So, what do we know? Let’s read the problem, carefully, and think about itl
“y = f(x) is a curve that always lies above the x-axis” ⇒ y > 0
“y = f(x) . . . never has a horizontal tangent” ⇒ f ’(x) ≠ 0 for any value of x
“f is differentiable everywhere” ⇒ f ’(x) exists for all values of x
“f is differentiable everywhere” ⇒ f(x) is defined everywhere and f(x) is continuous everywhere.
5
“For what value of y is the rate of change of y with respect to x eighty time the rate of
change of y with respect to x?” means we need to find the value(s) of y which satisfy:
5
Dx(y ) = 80
dy
dx
5
4
We know (from the Chain Rule) Dx(y ) = 5y Dx(y) = 5y
So our equation to solve is: 5y
Since we know that
4
4
4
dy
dx
dy
dy
= 80
dx
dx
dy
dy
≠ 0 [How do we know this?] we can divide each side by
.
dx
dx
4
5y = 80 → y = 16 → |y| = 2. → y = 2 or y = -2. HOWEVER, we were told y > 0, so
only
y=2
satisfies all of the conditions.
5