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Math 1431
Section 14839 M TH 4:00 PM-‐5:30 PM Online
Susan Wheeler
Office Hours: 5:30 - 6:15 pm M Th Online or by appointment
Wed 6:00 – 7:00 PM Online
Class webpage:
http://www.math.uh.edu/~swheeler/math1431.html The Exam Scheduler for Test 2 Should Open this Week
Stay Alert for This
Math 1431 Page 1 of 19 Section 2.3 !!
Section 2.3 – Differentiation Rules
We know how to take the derivative of polynomials and basic trigonometric functions. What if
the function is more complicated? How do we differentiate ( ) 25 1
4x
f xx
+=+
, ( ) ( )53g x x x= + , or
h x( ) = x2 sin x( ) ? In this section, we study some rules that will help us with these functions.
We will begin with the product rule. If a function is defined as the product of other functions, then we must use the product rule to differentiate that function.
Theorem 2.3.1: The Product Rule
If f and g are differentiable at x , then so is the product f ⋅ g . Moreover,
f ⋅ g( )' x( ) = f ' x( )g x( ) + f x( )g' x( )
This formula may be written as:
( )uv ' u' v uv'= + or ( )d du dvu v v u
dx dx dx⋅ = ⋅ + ⋅ .
This rule can be extended to the product of more than two functions:
( )uvw ' u' vw uv' w uvw'= + + or ( )d du dv dwu v w v w u w u v
dx dx dx dx⋅ ⋅ = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ .
Example 1: Find the derivative of h x( ) = x2 sin x( ) .
Solution: Since this function is the product of the functions ( ) 2f x x= and ( )g x sin x= , we
need to use the product rule:
ddx
h x( )⎡⎣ ⎤⎦ =ddx
x2( ) ⋅sin x + x2 ⋅ ddx
sin x( ) = 2x sin x + x2 cos x .
Math 1431 Page 2 of 19 Section 2.3 !!
Example 2: Find ( )2
2d
sec xdx
⎡ ⎤⎣ ⎦ .
Solution: Start with finding the first order derivative:
( ) ( ) ( )dsec x sec x tan x
dx=⎡ ⎤⎣ ⎦ .
Notice that the first derivative is the product of two functions; we need to use the product rule while differentiating it.
d2
dx2sec x( )⎡⎣ ⎤⎦ =
ddx
sec x( )tan x( )⎡⎣ ⎤⎦
= sec x( )tan x( )derivative of sec x( )! "## $##
⋅ tan x( ) + sec x( ) ⋅ sec2 x( )derivative of tan x( )!"# $#
= sec x( )tan2 x( ) + sec3 x( ) .
That is,
( ) ( ) ( ) ( )2
2 32
dsec x sec x tan x sec x
dx= +⎡ ⎤⎣ ⎦ .
Theorem 2.3.2: The Reciprocal Rule
If f is differentiable at x and ( ) 0f x ≠ , then so is the reciprocal 1f
. Moreover,
( )( )( ) 2
1 f ' xddx f x f x
⎡ ⎤= −⎢ ⎥
⎣ ⎦ ⎡ ⎤⎣ ⎦
Example 3: For ( ) 21
g xx x
=+
, find ( )3g' .
Math 1431 Page 4 of 19 Section 2.3 !!
Now, think about differentiating a rational function. If a function is the quotient of two functions, we need to use the quotient rule to differentiate that function.
Theorem 2.3.3: The Quotient Rule
If f and g are differentiable at x and ( ) 0g x ≠ , then the quotient f / g is differentiable at x
and
fg
⎛⎝⎜
⎞⎠⎟′
x( ) = f ' x( )g x( )− f x( )g' x( )g x( )⎡⎣ ⎤⎦
2
This formula may be written as:
2
'u u' v u v'v v
⋅ − ⋅⎛ ⎞ =⎜ ⎟⎝ ⎠ .
Since 1f fg g= ⋅ , the quotient rule can be obtained by using a combination of the product rule
and the reciprocal rule.
Example 5: Find the derivative of ( ) 25 1
4x
f xx
+=+
.
Solution: Since this function is the quotient of 5 1x + and 2 4x + , we need to use the quotient rule:
f ' x( ) =5x +1( )' x2 + 4( )− 5x +1( ) x2 + 4( )'
x2 + 4( )2
=5 x2 + 4( )− 5x +1( ) 2x( )
x2 + 4( )2
Math 1431 Page 6 of 19 Section 2.3 !!
Finally, how do we differentiate the composition of two functions? For instance, what is the
derivative of ( ) ( )52h x x x= + ? We see that expanding this expression and then taking the
derivative will be very tedious. Since this function is the composition of two functions:
( ) 5f x x= , ( ) 2g x x x= + and ( ) ( )( )h x f g x= ,
we can use the following rule.
Theorem 2.3.4: The Chain Rule
If g is differentiable at x and f is differentiable at g x( ) , then the composition f ! g is
differentiable at x . Moreover,
f ! g( )' x( ) = f ' g x( )( ) ⋅ g' x( ) .
This rule is one of the most important rules of differentiation. It helps with many complicated functions.
Example 7: Find the derivative of ( ) ( )52h x x x= + .
Solution: As stated earlier, h is the composition of two functions:
( ) 5f x x= , ( ) 2g x x x= + and h x( ) = f g x( )( ) = f ! g( ) x( ) .
That is, using the chain rule,
h' x( ) = f ! g( )' x( ) = f ' g x( )( ) ⋅ g' x( ) .
Here, ( ) 45f ' x x= , ( ) 2 1g' x x= + and ( )( ) ( )425f ' g x x x= + . Hence,
( ) ( )( ) ( ) ( ) ( )425 2 1h' x f ' g x g' x x x x= ⋅ = + + .
Math 1431 Page 7 of 19 Section 2.3 !!
Now that we know the derivative, we can evaluate it at given points. For example,
( ) ( ) ( )421 5 1 1 2 1 1 240h' = + ⋅ + = and ( ) ( ) ( )
420 5 0 0 2 0 1 0h' = ⋅ + ⋅ + = .
As seen in the previous example, the chain rule gives a somewhat “generalized” power rule:
Fact: If u is a differentiable function at x , and n is any real number other than 0, then
( ) 1n nd duu n u
dx dx−= ⋅ ⋅ .
Note that if n is a negative real number, this fact holds true for all x such that u x( ) ≠ 0 .
Example 8: For ( ) ( )506 11f x x= − , find ( )2f ' .
Solution: One can expand this function into a polynomial using the binomial theorem. However, this is not the best approach to solve this problem as it will be very difficult to keep track of those 51 terms. Using the chain rule is a much more efficient way.
( ) ( ) ( ) ( ) ( )49 49 4950 6 11 6 11 50 6 11 6 300 6 11df ' x x x x x
dx= ⋅ − ⋅ − = ⋅ − ⋅ = − .
Direct substitution gives:
( ) ( )49 492 300 6 2 11 300 1 300f ' = ⋅ − = ⋅ = .
Example 9: Find the derivative of ( )41
1x
h xx−⎛ ⎞= ⎜ ⎟+⎝ ⎠
at 0x = .
Solution: Use the chain rule:
( )31 14
1 1x d x
h' xx dx x− −⎛ ⎞ ⎛ ⎞= ⋅ ⋅⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
.
To find the derivative of 11
xx−+
, use the quotient rule: Math 1431 Page 10 of 19 Section 2.3 !
Hence,
f ' x( ) = g ! h( )' x( ) = g' h x( )( ) ⋅h' x( ) .
Here,
( ) 23 1h' x x= − , ( )g' x sin x= − and ( )( ) ( )3g' h x sin x x= − − .
The derivative is:
f ' x( ) = g' h x( )( ) ⋅h' x( ) = − sin x3 − x( ) ⋅ 3x2 −1( ) = − 3x2 −1( )sin x3 − x( ) ⋅ .
We can generalize this approach to all basic trigonometric functions.
The Chain Rule Applied to Six Trigonometric Functions
If u is a differentiable function of x , then
ddx
sinu( ) = cosu ⋅ dudx
ddx
cosu( ) = − sinu ⋅ dudx
ddx
tanu( ) = sec2 u ⋅ dudx
ddx
cot u( ) = −csc2 u ⋅ dudx
ddx
secu( ) = secu ⋅ tanu ⋅ dudx
ddx
cscu( ) = −cscu ⋅cot u ⋅ dudx
More Examples: 1. Find the derivative:
a. f x( ) = − 1
x2
b. f x( ) = x2 +2( )
x3
c. f x( ) = x3 + 3x
x2 −1
2. Find ( )0f ' given that h (0) = 3 and h’ (0) = 2 if f x( ) = h x( ) + x
h x( )
3. Find ddx
3 f (x)+ x2( )3( )
4. Find an equation for the tangent line to the graph of f x( ) = x2 − 10
x
at the point (-2, f(-2)).
5. Find x such that a) f ’(x) = 0, b) f ’ (x) > 0, c) f ’ (x) < 0
Given that f x( ) = 3x 4 − 4x3 − 2
6. 5. Find an equation for the normal line to the graph of f (x) = sin2 x at the
point π4, f π
4⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟ .
MORE examples:
7. Given f (x) = x3 and g(x) = sin(x)
If, h(x) = f ⋅g( ) x( ), find ′h x( ) .
8. If, h(x) = fg( ) x( ), find ′h x( ) find h '(x) .
9. If, h(x) = f g(x)( ), find ′h x( ) .
10. Given: f (2) = 7, f '(2) =1, f (5) = 4, f '(5) = 4,g(2) = 5, g '(2) = 3, g(5) =10, g'(5) = 6.
a) If ( ) ( )( )h x fg x= , find '(2)h .
b) If h(x) = f
g⎛⎝⎜
⎞⎠⎟
x( ) , find h '(2) .
Given: f (2) = 7, f '(2) =1, f (5) = 4, f '(5) = 4,g(2) = 5, g '(2) = 3, g(5) =10, g'(5) = 6.
c) If ( )3
( )h x f x= ⎡ ⎤⎣ ⎦ , find '(2)h . d) If h(x) = ( f ! g)(x) , find '(2)h .
Implicit Differentiation
Section 2.4 If we can’t solve, or can’t solve easily, for y as a function of x, use implicit differentiation to find the derivative of y. Discuss: + =2 2x y 1
“Rules” for implicit differentiation 1) Differentiate both sides of the equation with respect to x. 2) Collect all dy
dx (or y ’ ) terms.
3) Factor out dy
dx (or y ’ ) .
4) Solve for dy
dx (or y ’ ) .
If solving in terms of x, take the derivative as usual. If solving in terms of y, use the chain rule.
For the following, use implicit differentiation to find the derivative. 1. = 4y x 2. ( )3d x ydx
3. =2x y 5
4. + =3 22x y 8 5. ( )sin 2d ydx
6. ( )sin2d ydx
7. + − + =3 2y 2y 3y x 2
8. 9. sin cos =2 x y 1
2 3x 3xy y 10+ + =
10. Find the slope of the graph at the given point, then find the equation of the tangent line at (1, 1) for + =3 3x y 2xy.
11. Find the second derivative of − =2 2x y 16 in terms of y.