Section 3.6:The Chain Rule
1
The Chain Rule
If we have a function that is the composition of two (or more) functions, we use
the Chain Rule to differentiate it:
d
f (g(x)) = f 0 (g(x))g 0 (x)
dx
2
Homework Problems
1. The function h(x) = (6x2 + 7)−1/2 may be viewed as a composite function
h(x) = f (g(x)). Find f (x) and g(x).
Solution: g(x) = 6x2 + 7,
f (x) = x−1/2
2. The function Q(x) = sin4 (8x + 7) may be viewed as a composite function
Q(x) = f (g(h(x))). Find f (x), g(x) and h(x).
Solution: f (x) = x4 ,
g(x) = sin x
h(x) = 8x + 7
√
3. Find the derivative of the function y = 1 − 9x.
Solution: Using the Chain Rule, y 0 =
1
−9
(1 − 9x)−1/2 = √
2
2 1 − 9x
4. Find the derivative of the function y = tan(6x5 ).
Solution: Using the Chain Rule, y 0 = sec2 (6x5 )(30x4 )
5. Find the derivative of the function r = tan(6 − 3θ).
Solution: Using the Chain Rule, r0 = −3 sec2 (6 − 3θ)
√
6. Find the derivative of the function y = x tan(4 x) + 10.
√
Solution:
Using the Chain Rule and the Product Rule, y 0 = x(sec2 (4 x)(2x−1/2 )+
√
tan(4 x)
1
7. Let h(x) = f (g(x)) and p(x) = g(f (x)). Use the table to compute h0 (1)
and p0 (3)
x
1
f (x)
3
f 0 (x) −8
1
g(x)
g 0 (x) 8/9
2
1
−3
2
4/9
3
2
−2
4
5/9
4
4
−5
3
2/9
Solution Using the Chain Rule, we know that
h0 (1) = f 0 (g(1))g 0 (1) = f 0 (1)(8/9) = (−8)(8/9) = −64/9
p0 (3) = g 0 (f (3))f 0 (3) = g 0 (2)(−2) = (4/9)(−2) = −8/9
8. Find the derivative of the function y =
√
3 + cot2 x.
Solution: Using the Chain Rule (Twice),
y0 =
1
− csc2 x cot x
(3 + cot2 x)−1/2 (2 cot x)(− csc2 x) = √
2
3 + cot2 x
9. Find the derivative of the function y = cos9 (sin(7x)).
Solution: Using the Chain Rule (Four Times),
y 0 = 9 cos8 (sin(7x))(− sin(sin(7x)))(cos(7x))(7) = −63 cos8 (sin(7x)) sin(sin(7x)) cos(7x)
10. Find the derivative of the function f (g(7x4 )).
Solution: Using the Chain Rule (Twice),
d
= f 0 (g(7x4 ))g 0 (7x4 )(28x3 )
dx
2
11. Find the second derivative of the function y = 4x cos(x2 ).
Solution: Using the Chain Rule and the Product Rule,
y 0 = 4x(− sin(x2 )(2x)) + 4 cos(x2 ) = −8x2 sin(x2 ) + 4 cos(x2 )
Which gives that
y 00 (x) = −8x2 (cos(x2 ))(2x) − 16x sin(x2 ) + 4(− sin(x2 ))(2x)
= −16x3 cos(x2 ) − 24x sin(x2 )
12. Find the equations of the tangent line to the graph of y =
(x2
the point (−2, −1/2)
Solution Using the Quotient and Chain Rules,
y0 =
=
(x2 − 6)4 (3x2 ) − (x3 )(4(x2 − 6)3 )(2x)
(x2 − 6)8
3x2 (x2 − 6)4 − 8x4 (x2 − 6)3
3x2 (x − 6) − 8x4
=
(x2 − 6)8
(x2 − 6)5
So, the slope of the tangent line at the point x = 2 is
y 0 (1) =
3(2)2 (4 − 6) − 8(2)4
12(−2) − 128
19
=
=
(4 − 6)5
−45
4
So the equation of the tangent line is
y + 1/2 = 19/4(x + 2)
or
y = 19/4x + 9
3
x3
at
− 6)4
13. Let h(x) = f (g(x)). Suppose the equation of the tangent line to the graph
of g at the point (3, 2) is y = 4x − 10 and the equation of the tangent line
to the graph of f at (2, 9) is y = −3x + 15. a) Calculate h(3) and h0 (3).
b) Find the equation of the tangent line to the graph of h at the point x = 3
Solution
We know that h(3) = f (g(3)) where, by plugging 3 into the equation of
the tangent line of g, g(3) = 2. Using the same process, we know that
f (2) = 9. So,
h(3) = f (g(3)) = f (2) = 9
We know that h0 (3) = f 0 (g(3))g 0 (3), where g 0 (3) = 4, which is the slope
of the tangent line at x = 3. Also, f 0 (2) = −3 (Using the same process).
So,
h0 (3) = f 0 (g(3))g 0 (3) = f 0 (2)(4) = (−3)(4) = −12
Using this information, the equation of the tangent line of the graph of h
at the point x = 3 is
y − 9 = −12(x − 3)
or
y = −12x + 45
14. Let f (2) = 1 and f 0 (2) = −5. Let g(x) = sin(πf (x)). Evaluate g 0 (2).
Solution
g 0 (2) = cos(πf (2))(πf 0 (2)) = cos(π)(−5π) = 5π
4