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Derivatives
Definition and Notation
If y = f (^) ( x )then the derivative is defined to be (^) ( ) ( ) ( ) 0 lim h
f x h f x f x Æ h
If y = f (^) ( x )then all of the following are
equivalent notations for the derivative.
( ) ( ( )) ( )
df dy d f x y f x Df x dx dx dx
If y = f (^) ( x )all of the following are equivalent notations for derivative evaluated at x = a.
( ) (^) x a ( ) x a x a
df dy f a y Df a = (^) dx (^) = dx =
Interpretation of the Derivative
If y = f (^) ( x )then,
- m = f ¢( a )is the slope of the tangent line to y = f (^) ( x )at x = a and the equation of the tangent line at x = a is given by y = f (^) ( a (^) ) + f ¢( a (^) )( x - a ). 2. f ¢^ ( a )is the instantaneous rate of change of f (^) ( x (^) )at x = a. 3. If f (^) ( x (^) )is the position of an object at time x then f ¢^ ( a )is the velocity of the object at x = a.
Basic Properties and Formulas
If f (^) ( x (^) )and g (^) ( x (^) )are differentiable functions (the derivative exists), c and n are any real numbers,
- ( c f ) ¢^ = c f ¢( x )
- ( f ± g ) ¢^ = f ¢^ ( x ) ± g ¢( x )
- ( f g )¢^ = f ¢^ g + f g ¢ – Product Rule
f f g f g g g
Ê ˆ^ ¢ ¢^ - ¢
Á ˜ =
Ë ¯
- Quotient Rule 5. ( ) 0
d c dx
- (^) ( n^ ) n^1
d x n x dx
= - – Power Rule
- (^) ( ( ( ))) ( ( )) ( )
d f g x f g x g x dx
= ¢^ ¢
This is the Chain Rule
Common Derivatives
( ) 1
d x dx
( sin^ ) cos
d x x dx
( cos^ ) sin
d x x dx
( tan^ ) sec^2
d x x dx
( sec^ ) sec^ tan
d x x x dx
( csc^ ) csc^ cot
d x x x dx
( cot^ ) csc^2
d x x dx
( ) 1 2
sin 1
d x dx (^) x
( ) 1 2
cos 1
d x dx (^) x
( ) 1 2
tan 1
d x dx x
( ) ln(^ )
d (^) a x (^) a x a dx
( )
d x x dx
e = e
( ( ))
ln , 0 d x x dx x
( )
ln , 0 d x x dx x
= π
( ( ))
log , 0 ln a
d x x dx x a
Chain Rule Variants The chain rule applied to some specific functions.
d n n 1 f x n f x f x dx
ÈÎ ˘˚ = ÈÎ ˘˚^ ¢
2. ( f^ (^ x )^ ) ( ) f^ (^ x )
d f x dx
e = ¢ e
ln
d f^ x f x dx f x
ÈÎ ˘˚ =
4. ( sin ( ) ) ( ) cos ( )
d f x f x f x dx
ÈÎ ˘˚ = ¢ ÈÎ ˘˚
5. ( cos ( ) ) ( ) sin ( )
d f x f x f x dx
ÈÎ ˘˚ = - ¢ ÈÎ ˘˚
6. ( tan ( ) ) ( ) sec^2 ( )
d f x f x f x dx
ÈÎ ˘˚ = ¢ ÈÎ ˘˚
7. ( sec [ f ( ) x ]) f ( ) x sec[ f ( ) x ] tan[ f ( ) x ]
d dx
1 tan (^2) 1
d f^ x f x dx (^) f x
- È ˘ = ¢
Î ˚
+ ÈÎ ˘˚
Higher Order Derivatives The Second Derivative is denoted as
( ) (^ )^ ( )
2 2 2
d f f x f x dx
¢¢ = = and is defined as
f ¢¢ ( x ) = ( f ¢( x ))¢, i.e. the derivative of the
first derivative, f ¢ ( x ).
The nth^ Derivative is denoted as
n n n
d f f x dx
= and is defined as
f n^ x = f n -^1 x ¢, i.e. the derivative of
the ( n -1)st^ derivative, f (^ n -^1 )^ ( x ).
Implicit Differentiation
Find y ¢^ if e^2^ x^ -^9 y^ + x y^3 2^ = sin ( y )+ 11 x. Remember y = y x ( )here, so products/quotients of x and y
will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to differentiate as normal and every time you differentiate a y you tack on a y ¢^ (from the chain rule).
After differentiating solve for y ¢.
( (^ ))
2 9 2 2 3 2 9 2 2 2 9 2 9 2 2 3 3 2 9 3 2 9 2 9 2 2
2 9 3 2 cos 11 11 2 3 2 9 3 2 cos 11 2 9 cos 2 9 cos 11 2 3
x y x y x y x y x y x y x y
y x y x y y y y x y y x y x y y y y y x y y x y y y x y
e e e e e e e
Increasing/Decreasing – Concave Up/Concave Down Critical Points
x = c is a critical point of f ( x )provided either
1. f ¢ ( ) c = 0 or 2. f ¢ ( ) c doesn’t exist.
Increasing/Decreasing
1. If f ¢^ ( x ) > 0 for all x in an interval I then
f ( x )is increasing on the interval I.
2. If f ¢^ ( x ) < 0 for all x in an interval I then
f ( x )is decreasing on the interval I.
3. If f ¢^ ( x ) = 0 for all x in an interval I then
f ( x )is constant on the interval I.
Concave Up/Concave Down
1. If f ¢¢^ ( x ) > 0 for all x in an interval I then
f ( x )is concave up on the interval I.
2. If f ¢¢ ( x ) < 0 for all x in an interval I then
f ( x )is concave down on the interval I.
Inflection Points
x = c is a inflection point of f ( x )if the
concavity changes at x = c.
Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to t using implicit differentiation ( i.e. add on a derivative every time you differentiate a function of t ). Plug in known quantities and solve for the unknown quantity. Ex. A 15 foot ladder is resting against a wall. The bottom is initially 10 ft away and is being pushed towards the wall at 14 ft/sec. How fast
is the top moving after 12 sec?
x ¢^ is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x^2^ + y^2 = 152 fi 2 x x ¢ + 2 y y ¢= 0
After 12 sec we have x = 10 - (^12) ( 14 )= 7 and
so y = 152 - 7 2 = 176. Plug in and solve
for y ¢^.
( 14 )
7 176 0 ft/sec 4 176
Ex. Two people are 50 ft apart when one starts walking north. The angle q changes at 0.01 rad/min. At what rate is the distance between them changing when q = 0.5rad?
We have q ¢^ = 0.01rad/min. and want to find x ¢^. We can use various trig fcns but easiest is,
sec sec tan 50 50
x x q q q q
= fi ¢=
We know q = 0.05so plug in q ¢^ and solve.
sec 0.5 tan 0.5 ( ) ( )( 0.01) 50 0.3112 ft/sec
x
x
Remember to have calculator in radians!
Optimization Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for one of the two variables and plug into first equation. Find critical points of equation in range of variables and verify that they are min/max as needed. Ex. We’re enclosing a rectangular field with 500 ft of fence material and one side of the field is a building. Determine dimensions that will maximize the enclosed area.
Maximize A = xy subject to constraint of
x + 2 y = 500. Solve constraint for x and plug
into area.
( ) 2
A y y x y y y
= - fi = -
Differentiate and find critical point(s). A ¢^ = 500 - 4 y fi y = 125 By 2nd^ deriv. test this is a rel. max. and so is the answer we’re after. Finally, find x.
x = 500 - 2 125 ( ) = 250
The dimensions are then 250 x 125.
Ex. Determine point(s) on y = x^2 + 1 that are closest to (0,2).
Minimize (^) ( ) ( ) 2 2 2 f = d = x - 0 + y - 2 and the constraint is y = x^2 + 1. Solve constraint for x^2 and plug into the function. ( )
( )
2 2 2 (^2 )
x y f x y
y y y y
= - fi = + -
= - + - = - + Differentiate and find critical point(s). 3 f ¢^ = 2 y - 3 fi y = 2 By the 2nd^ derivative test this is a rel. min. and so all we need to do is find x value(s). (^2 3 1 ) x = 2 - 1 = 2 fi x = ± 2
The 2 points are then (^) ( 12 , (^32) )and (^) ( -^12 ,^32 )
