Derivatives of Trigonometry
Derivative Rules Notation: y '
f '( x)
Rules: f ( x) = c
dy dx
d dx
Dx
f '( x) =
f ( x ) = xn
f '( x) =
f '( x) =
f ( x) = a
x
f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = log a x
f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = loga x
f '( x) =
f ( x ) = sin x
f '( x) =
f ( x ) = cos x
f '( x) =
f ( x ) = sec x
f '( x) =
f ( x ) = csc x
f '( x) =
f ( x ) = tan x
f '( x) =
f ( x ) = cot x
f '( x) =
f ( x ) = sin −1 x
f '( x) =
f ( x ) = cos−1 x
f '( x) =
f ( x ) = sec−1 x
f '( x) =
f ( x ) = csc−1 x
f '( x) =
f ( x ) = tan −1 x
f '( x) =
f ( x ) = cot −1 x
f '( x) =
f ( x) = e
x
d ( f ( x ) ± g ( x )) = dx
d (cf ( x )) = dx
© 2016 Study Edge®
Derivative Applications Part 1 f ( x) = c f '( x) =
f ( x ) = xn
d ( f ( x ) ± g ( x )) = dx
d (cf ( x )) = dx
f '( x) =
A) Find the derivative at a given x-value: Easiest case. Take the derivative, plug in the x-value. S1: Find f ' ( −1) for f ( x ) = 4 x3 − 2 x + 1
B) Equations of tangent/normal lines: What do you need for a line? Equation: S2: Find the equation of the tangent line to f ( x ) = 6 x − 4 x at x = 1
© 2016 Study Edge®
C) Horizontal Tangents: S3: Find the x-values of the horizontal tangent lines to f ( x ) = 3x 4 − 4 x3 + 4
D) Finding slopes: S4: Find all x values where the slope of the tangent line to f ( x ) = 2 x 2 + 5x + 5 is perpendicular to the line x + 3 y = 13 .
© 2016 Study Edge®
Derivative Applications Part 2 E) Average vs. Instantaneous Average = slope Instantaneous = Derivative S1: The number of drinks consumed on a Friday night downtown is given by D ( x ) = 3x2 − 2 x + 5 , where D is the number of drinks and x is the number of hours since the bar opens. What is the average number of drinks consumed in the first 3 hours, and at what rate are you drinking by the end of the third hour (x = 3).
F)
Position
–
Velocity
–
Acceleration
S2: The position of a particle is given by s (t ) = 4t 3 − 6t 2 + 5 , where t is in seconds and s (t ) is in feet. a) Find the velocity and acceleration equations. b) When is the particle at rest?
S3: An object is thrown in the air follows the equation h (t ) = −16t 2 + 80t + 96 where t is in seconds and
h (t ) is in feet. a) Find the maximum height. b) Find the velocity as the object hits the ground.
© 2016 Study Edge®
Product Rule: h( x) = f ( x) g ( x) , then h '( x) = f ( x) g '( x) + g ( x) f '( x) Derivative of [first * second] = [First * Derivative of Second] + [Second * Derivative of First] S1: Find f ' ( x ) for f ( x ) = ( 2 x + 1)(3x − 7 ) .
© 2016 Study Edge®
Quotient Rule:
h( x ) =
f ( x) , g ( x)
then
top f ( x) = bottom f ( x) =
high low
S1: Find f ' ( x ) for f ( x ) =
h '( x) =
g ( x) f '( x) − f ( x) g '( x)
( g ( x) ) bottom *
f '( x) =
f '( x) =
2
d d (top) − top * (bottom) dx dx 2 (bottom)
LdH − HdL L2
3x 2 − 5 . x
© 2016 Study Edge®
Chain Rule: h( x) = f ( g ( x))
then
h( x) = f '( g ( x)) g '( x)
2
S1: Find f ' ( x ) for f ( x ) = ( 4x + 5) .
© 2016 Study Edge®
Study Edge Tip It helps to ask yourself do I have a function where (a) two equations are being muliplied then use the Product Rule (b) two equations are being divided then use the Quotient Rule (c) everything is surrounded by a function then use the Chain Rule Polynomials/Rational Functions 4
S1: f ( x) = ( 2 x − 5 ) ( 2 − x 2 ) find f '( x ) . 3
4
S2: Find the equation of the normal line to f ( x ) = −2 x (3x + 2) at x = −1 .
© 2016 Study Edge®
S3: Find the x-values of the horizontal tangent lines to f ( x ) =
x3 x−4
© 2016 Study Edge®
Exponential Functions f ( x ) = ex f '( x) =
f ( x) = ax
f '( x) =
3
S1: f ( x ) = e4 x +5 x−7 find f '( x ) .
S2: f ( x ) =
e5 x find f ' ( 0) e3 x + 2
© 2016 Study Edge®
S3: Find the x-values of the horizontal tangent lines to f ( x ) = x3e5x .
S4: Find the equation of the tangent line to f ( x ) = e2 x + 3 at x = 0 .
© 2016 Study Edge®
Derivatives for general functions Product Rule: h( x ) = f ( x ) g ( x ) ,
f ( x) , g ( x)
Quotient Rule:
h( x ) =
Chain Rule:
h( x) = f ( g ( x))
then
h '( x) = f ( x) g '( x) + g ( x) f '( x)
then
h '( x) =
then
h( x) = f '( g ( x)) g '( x)
g ( x) f '( x) − f ( x) g '( x)
( g ( x) )
2
S1: Given that f (1) = 4, f '(1) = −1, g '(1) = 2, f '(3) = 6, g '(3) = −2, g (1) = 3, and h( x) = ( f o g )( x) , find h '(1) .
S2: Let g ( x) = xf x2 + 1 , f (2) = 7 and f '(2) = −3 Find g '(1) .
(
)
© 2016 Study Edge®
Trigonometric Functions
f ( x ) = sin x
f '( x) =
f ( x ) = cos x
f '( x) =
f ( x ) = sec x
f '( x) =
f ( x ) = csc x
f '( x) =
f ( x ) = tan x
f '( x) =
f ( x ) = cot x
f '( x) =
© 2016 Study Edge®
S1: f ( x ) = cos2 (sin 3x ) find f '( x ) .
S2: f ( x ) =
cos x find the x-values of the horizontal tangent lines on the interval [0, 2π ]. 1 + sin x
© 2016 Study Edge®
Implicit Differentiation dy Keep in mind that , or y’, means derivative of y with respect to x. dx S1: Find
dy for y + x 4 y 2 = x 2 − 4 y 3 + 5 . dx
4
S2: What is the slope of the tangent line of 4 x3 + ( x + y ) = 3 y + 5 at (1,0 )
© 2016 Study Edge®
Inverse Trigonometric Functions f ( x ) = sin −1 x f '( x) =
f ( x ) = cos−1 x
f '( x) =
f ( x ) = sec−1 x
f '( x) =
f ( x ) = csc−1 x
f '( x) =
f ( x ) = tan −1 x
f '( x) =
f ( x ) = cot −1 x
f '( x) =
⎛ x⎞ S1: f ( x) = sin −1 ⎜ ⎟ ⎝4⎠
S2: f ( x) = arctan 3x4
( )
© 2016 Study Edge®
Logarithmic Functions f ( x ) = ln x f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = log a x
f '( x) =
f ( x ) = loga x
f '( x) =
S1: f ( x ) = ln x 2 find f '( x ) .
2
S2: f ( x ) = ( ln x ) find f '( x ) .
⎛ e4 x ⎞ S3: f ( x ) = ln ⎜ 2 x ⎟ find f '( x ) . ⎝ e +5⎠
S4: f ( x ) = x ln x find f '( x ) .
© 2016 Study Edge®
Logarithmic Differentiation For complicated functions (they will specify to use) OR If you have a function raised to a function such as f ( x ) = x x+1 , f ( x ) = xln x , f ( x ) = ( x + 5)
x
Step 1: Make sure to set f(x) = y, take the ‘ln’ of both sides, and expand. Step 2: Take the derivative. Step 3: Multiply by y at the end. S1: Find the derivative of f ( x) =
cos 2 x ln x e3 x
S2: Find the derivative of f ( x) = xln x .
© 2016 Study Edge®
Related Rates Key words: per hour, per minute, respect to time Step 1: Find the function that relates everything. Step 2: Write down all the variables, and their rates of change, fill in what you have. Step 3: Take the derivative with respect to time. Revenue: R( x) = xp
Profit: P( x) = R( x) − C ( x)
Distance: d = ( x2 − x1 )2 + ( y2 − y1 )2 Volumes:
Areas:
Cylinder V = π r 2 h
Average Cost:
C ( x) x
Pythagorean Theorem: a 2 + b 2 = c 2 1 Cone V = π r 2 h 3
4 Sphere V = π r 3 3
Cube V = s 3
Circle
1 Triangle: A = bh 2
A = π r2
C ( x) =
S1: A big ladder falls so the base is sliding at a rate of 20 ft/sec when it is 300 ft from the base of a tower, how fast is the top sliding when it is at a height of 400 ft?
© 2016 Study Edge®
S2: Sand is falling to the ground and making a conical pile so the height is increasing by 11 cm/sec. Find how fast the volume is increasing the instance the height is 2 cm and the radius is 6cm.
© 2016 Study Edge®
© 2016 Study Edge®
f '( x)
Rules: f ( x) = c
dy dx
d dx
Dx
f '( x) =
f ( x ) = xn
f '( x) =
f '( x) =
f ( x) = a
x
f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = log a x
f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = loga x
f '( x) =
f ( x ) = sin x
f '( x) =
f ( x ) = cos x
f '( x) =
f ( x ) = sec x
f '( x) =
f ( x ) = csc x
f '( x) =
f ( x ) = tan x
f '( x) =
f ( x ) = cot x
f '( x) =
f ( x ) = sin −1 x
f '( x) =
f ( x ) = cos−1 x
f '( x) =
f ( x ) = sec−1 x
f '( x) =
f ( x ) = csc−1 x
f '( x) =
f ( x ) = tan −1 x
f '( x) =
f ( x ) = cot −1 x
f '( x) =
f ( x) = e
x
d ( f ( x ) ± g ( x )) = dx
d (cf ( x )) = dx
© 2016 Study Edge®
Derivative Applications Part 1 f ( x) = c f '( x) =
f ( x ) = xn
d ( f ( x ) ± g ( x )) = dx
d (cf ( x )) = dx
f '( x) =
A) Find the derivative at a given x-value: Easiest case. Take the derivative, plug in the x-value. S1: Find f ' ( −1) for f ( x ) = 4 x3 − 2 x + 1
B) Equations of tangent/normal lines: What do you need for a line? Equation: S2: Find the equation of the tangent line to f ( x ) = 6 x − 4 x at x = 1
© 2016 Study Edge®
C) Horizontal Tangents: S3: Find the x-values of the horizontal tangent lines to f ( x ) = 3x 4 − 4 x3 + 4
D) Finding slopes: S4: Find all x values where the slope of the tangent line to f ( x ) = 2 x 2 + 5x + 5 is perpendicular to the line x + 3 y = 13 .
© 2016 Study Edge®
Derivative Applications Part 2 E) Average vs. Instantaneous Average = slope Instantaneous = Derivative S1: The number of drinks consumed on a Friday night downtown is given by D ( x ) = 3x2 − 2 x + 5 , where D is the number of drinks and x is the number of hours since the bar opens. What is the average number of drinks consumed in the first 3 hours, and at what rate are you drinking by the end of the third hour (x = 3).
F)
Position
–
Velocity
–
Acceleration
S2: The position of a particle is given by s (t ) = 4t 3 − 6t 2 + 5 , where t is in seconds and s (t ) is in feet. a) Find the velocity and acceleration equations. b) When is the particle at rest?
S3: An object is thrown in the air follows the equation h (t ) = −16t 2 + 80t + 96 where t is in seconds and
h (t ) is in feet. a) Find the maximum height. b) Find the velocity as the object hits the ground.
© 2016 Study Edge®
Product Rule: h( x) = f ( x) g ( x) , then h '( x) = f ( x) g '( x) + g ( x) f '( x) Derivative of [first * second] = [First * Derivative of Second] + [Second * Derivative of First] S1: Find f ' ( x ) for f ( x ) = ( 2 x + 1)(3x − 7 ) .
© 2016 Study Edge®
Quotient Rule:
h( x ) =
f ( x) , g ( x)
then
top f ( x) = bottom f ( x) =
high low
S1: Find f ' ( x ) for f ( x ) =
h '( x) =
g ( x) f '( x) − f ( x) g '( x)
( g ( x) ) bottom *
f '( x) =
f '( x) =
2
d d (top) − top * (bottom) dx dx 2 (bottom)
LdH − HdL L2
3x 2 − 5 . x
© 2016 Study Edge®
Chain Rule: h( x) = f ( g ( x))
then
h( x) = f '( g ( x)) g '( x)
2
S1: Find f ' ( x ) for f ( x ) = ( 4x + 5) .
© 2016 Study Edge®
Study Edge Tip It helps to ask yourself do I have a function where (a) two equations are being muliplied then use the Product Rule (b) two equations are being divided then use the Quotient Rule (c) everything is surrounded by a function then use the Chain Rule Polynomials/Rational Functions 4
S1: f ( x) = ( 2 x − 5 ) ( 2 − x 2 ) find f '( x ) . 3
4
S2: Find the equation of the normal line to f ( x ) = −2 x (3x + 2) at x = −1 .
© 2016 Study Edge®
S3: Find the x-values of the horizontal tangent lines to f ( x ) =
x3 x−4
© 2016 Study Edge®
Exponential Functions f ( x ) = ex f '( x) =
f ( x) = ax
f '( x) =
3
S1: f ( x ) = e4 x +5 x−7 find f '( x ) .
S2: f ( x ) =
e5 x find f ' ( 0) e3 x + 2
© 2016 Study Edge®
S3: Find the x-values of the horizontal tangent lines to f ( x ) = x3e5x .
S4: Find the equation of the tangent line to f ( x ) = e2 x + 3 at x = 0 .
© 2016 Study Edge®
Derivatives for general functions Product Rule: h( x ) = f ( x ) g ( x ) ,
f ( x) , g ( x)
Quotient Rule:
h( x ) =
Chain Rule:
h( x) = f ( g ( x))
then
h '( x) = f ( x) g '( x) + g ( x) f '( x)
then
h '( x) =
then
h( x) = f '( g ( x)) g '( x)
g ( x) f '( x) − f ( x) g '( x)
( g ( x) )
2
S1: Given that f (1) = 4, f '(1) = −1, g '(1) = 2, f '(3) = 6, g '(3) = −2, g (1) = 3, and h( x) = ( f o g )( x) , find h '(1) .
S2: Let g ( x) = xf x2 + 1 , f (2) = 7 and f '(2) = −3 Find g '(1) .
(
)
© 2016 Study Edge®
Trigonometric Functions
f ( x ) = sin x
f '( x) =
f ( x ) = cos x
f '( x) =
f ( x ) = sec x
f '( x) =
f ( x ) = csc x
f '( x) =
f ( x ) = tan x
f '( x) =
f ( x ) = cot x
f '( x) =
© 2016 Study Edge®
S1: f ( x ) = cos2 (sin 3x ) find f '( x ) .
S2: f ( x ) =
cos x find the x-values of the horizontal tangent lines on the interval [0, 2π ]. 1 + sin x
© 2016 Study Edge®
Implicit Differentiation dy Keep in mind that , or y’, means derivative of y with respect to x. dx S1: Find
dy for y + x 4 y 2 = x 2 − 4 y 3 + 5 . dx
4
S2: What is the slope of the tangent line of 4 x3 + ( x + y ) = 3 y + 5 at (1,0 )
© 2016 Study Edge®
Inverse Trigonometric Functions f ( x ) = sin −1 x f '( x) =
f ( x ) = cos−1 x
f '( x) =
f ( x ) = sec−1 x
f '( x) =
f ( x ) = csc−1 x
f '( x) =
f ( x ) = tan −1 x
f '( x) =
f ( x ) = cot −1 x
f '( x) =
⎛ x⎞ S1: f ( x) = sin −1 ⎜ ⎟ ⎝4⎠
S2: f ( x) = arctan 3x4
( )
© 2016 Study Edge®
Logarithmic Functions f ( x ) = ln x f '( x) =
f ( x ) = ln x
f '( x) =
f ( x ) = log a x
f '( x) =
f ( x ) = loga x
f '( x) =
S1: f ( x ) = ln x 2 find f '( x ) .
2
S2: f ( x ) = ( ln x ) find f '( x ) .
⎛ e4 x ⎞ S3: f ( x ) = ln ⎜ 2 x ⎟ find f '( x ) . ⎝ e +5⎠
S4: f ( x ) = x ln x find f '( x ) .
© 2016 Study Edge®
Logarithmic Differentiation For complicated functions (they will specify to use) OR If you have a function raised to a function such as f ( x ) = x x+1 , f ( x ) = xln x , f ( x ) = ( x + 5)
x
Step 1: Make sure to set f(x) = y, take the ‘ln’ of both sides, and expand. Step 2: Take the derivative. Step 3: Multiply by y at the end. S1: Find the derivative of f ( x) =
cos 2 x ln x e3 x
S2: Find the derivative of f ( x) = xln x .
© 2016 Study Edge®
Related Rates Key words: per hour, per minute, respect to time Step 1: Find the function that relates everything. Step 2: Write down all the variables, and their rates of change, fill in what you have. Step 3: Take the derivative with respect to time. Revenue: R( x) = xp
Profit: P( x) = R( x) − C ( x)
Distance: d = ( x2 − x1 )2 + ( y2 − y1 )2 Volumes:
Areas:
Cylinder V = π r 2 h
Average Cost:
C ( x) x
Pythagorean Theorem: a 2 + b 2 = c 2 1 Cone V = π r 2 h 3
4 Sphere V = π r 3 3
Cube V = s 3
Circle
1 Triangle: A = bh 2
A = π r2
C ( x) =
S1: A big ladder falls so the base is sliding at a rate of 20 ft/sec when it is 300 ft from the base of a tower, how fast is the top sliding when it is at a height of 400 ft?
© 2016 Study Edge®
S2: Sand is falling to the ground and making a conical pile so the height is increasing by 11 cm/sec. Find how fast the volume is increasing the instance the height is 2 cm and the radius is 6cm.
© 2016 Study Edge®
© 2016 Study Edge®
