rate of change
dependent" per "independent
average rate of change on the interval [a,b] is represented by...
f(b) - f(a) / b - a
find the average rate of change of f(x) = ln 3x over the interval 1<=x<=4
f(4) - f(1) / (4 - 1) = (ln(3
4) - ln(3
1))/3 = 0.462
find the average rate of change over the interval 2<=x<=30.
x: 0, 2, 7, 30
f(x): 3, -2, 5, 7
(7 - -2)/(30 - 2) = 9/28
average rate of change:
the following quotients express the average rate of change of a function over an interval.
(f(a+h) - f(a))/((a+h) - a)
or
(f(x) - f(a))/ (x - a)
this is also the slope of the secant line
instantaneous rate of change:
the following limits express the instantaneous rate of change of a function at x = a.
lim ((f(a+h) - f(a))/h) h?0
or
lim ((f(x) - f(a))/(x-a)) x?a
this is also the slope of the tangent line at x = a
find the instantaneous rate of change of f(x) = x - x^2 at x = -1
do both ways
3 for h?0
3 for x?-1
identify the function we are working with. then identify the x-value for the instantaneous rate of change (slope of the tangent line at a point).
1. lim ((5ln(2/(4+h)) - 5ln(1/2))/h) h?0
2. lim ((sinx - 1)/(x - ?/2)) x??/2
1. function: f(x) = 5ln(2/x); instantaneous rate at x = 4
2. function: f(x) = sinx; instantaneous rate at x = ?/2
find the average rate of change of each function on the given interval. use the appropriate units if necessary.
1. f(x) = x^2 - 2; [-1,3]
2. A(t) = 2^t; [2,4]; t represents years, A represents dollars
3. h(m) = tan(m) + 4; [?/4, 3?/4]; h represents hair,
1. 2
2. $6/year
3. -1.273 hair/months
4. 0.251
5. 0.460
use the following table to find the average rate of change on the given interval.
t (minutes): 0, 3, 4, 12, 13
s(t) (feet): -2, 4, -7, 5, 10
6. [3,13]
7. 0<=x<=12
8. [3,4]
6. 0.600 ft/min
7. 7/12 ft/min
8. -11 ft/min
use the following graph to find the average rate of change on the given interval.
9, 10, and 11
look at page 3 in the packet
the graphs of f and g are given below. for each function, find the average rate of change on the given interval
12, 13, and 14
look at page 3 in the packet
for compositions: f(g(x)), find the g(x) first and then use those numbers to plug into f(x) and find those points
for addition h(x) = f(x) + g(x): solve for h(x) for both numbers on the interval by adding each f(x) and g(x) fo
find the instantaneous rate of change of each function at the given x-value. use the form lim ((f(a+h) - f(a))/h) h?0
15. f(x) = x^2 - x at x = -1
16. f(x) = ?x at x = 5
17. f(x) = 1/x at x = 2
15. -3
16. 1/(2?5)
17. -1/4
find the instantaneous rate of change of each function at the given x-value. use the form lim ((f(x) - f(a))/(x-a))
18. f(x) = 2x^2 + 3x - 4 at x = -3
19. f(x) = ?(3x) at x = 7
20. f(x) = 1/(5x) at x = -2
18. -9
19. 3/(2?21)
20. -1/20
each limit represents the instantaneous rate of change of a function. identify the original function, and the x-value of the instantaneous rate of change.
21 through 28
look at page 4 of the packet
this derivative is an expression that calculates the...
instantaneous rate of change (slope of the tangent line) of a function at any given x-value. in other words, it gives us the slope of the function at a point
f(x) = x^2
f'(x) =
f'(1) =
f'(2) =
f'(-2) =
1. if f'(x) = (5/x) - x, find f'(2) and explain the meaning
2. if f(x) represents how many meters you have run and x represents the minutes, describe in full sentences the following:
f(8) = 1,500
f'(3) = 161
2x
2
4
-4
1. f'(2) = 1/2; the slope of the tangent line of f(x) at x = 2 is 1/2
2. after 8 minutes, I ran 1,500 meters
at the 3rd minute, I was running 161 meters per minute
notation for the derivative:
Lagrange:
Leibniz:
f'(x)
dy/dx = the derivative of y with respect to x
definition of the derivative
this limit gives an expression that calculates the instantaneous rate of change (slope of the tangent line) of f(x) at any given x-value
f'(x) = lim ((f(x+h) - f(x))/h) h?0
find the derivative using the definition of the derivative (limits).
3. f(x) = 2x^2 - 7x + 1
4. y = 1/(x^2)
3. f'(x) = 4x - 7
4. dy/dx = -2/(x^3)
equation of the tangent line:
the line tangent to the curve of f(x) at x = a can be represented in point slope form:
y - f(a) = f'(a)(x - a)
5. if we know h(5) = -2, and the derivative of h is given by h'(x) = (x^3 - 2) / x, write an equation for the line tangent to the graph of h at x = 5.
y + 2 = (123/5)(x - 5)
6. the graph of f'(x), the derivative of f, is shown at the right. if f(2) = 7, write an equation of the line tangent to the graph of f at (2,7)
y - 7 = 3(x - 2)
(2,3) is the point on the graph f'(x)
find the derivative using limits. if the equation is given as y =, use Leibniz Notation: dy/dx. if the equation is given as f(x) =, use Lagrange Notation: f'(x).
1. f(x) = 7 - 6x
2. y = 5x^2 - x
3. y = ?(5x + 2)
4. f(x) = 1/(x - 2)
1. f'(x) = -6
2. dy/dx = 10x - 1
3. dy/dx = 5/(2?(5x + 2))
4. f'(x) = -1/((x - 2)^2)
for each problem, use the information given to identify the meaning of the two equations in the context of the problem. write in full sentences.
5. C is the number of championships Sully has won while coaching basketball. t is the number of years since 20
5. In 2014, Sully had won 3 basketball championships. At year 12 of coaching basketball, Sully is winning 0.4 championships per year.
6. At 7:30 am, I am 1.2 miles from home. At 7:30 am, I am walking 11 miles per hour back home.
7. If Mr. Kelly had 7 kids
for each problem, create an equation of the tangent line of f at the given point. leave in point-slope.
9. f(7) = 5 and f'(7) = -2
10. f(-2) = 3 and f'(-2) = 4
11. f(x) = 3x^2 + 2x; f'(x) = 6x + 2; x = -2
12. f(x) = 10?(6x + 1); f'(x) = 30/?(6x + 1); x =
9. y - 5 = -2(x - 7)
10. y - 3 = 4(x + 2)
11. y - 8 = -10(x + 2)
12. y - 50 = 6(x - 4)
13. y = -2(x - ?/4)
14. y - ?3 = 4(x - ?/3)
estimating the derivative with a calculator:
MATH
[8]: nDeriv(
Estimating the Derivative with a CALCULATOR:
1. if f(x) = sin ?x, find f'(2)
2. If f(x) = ln(1/(5-x)), find f'(1.3)
3. Write the equation of the line tangent to y = ?x/(x^3 + 1) at x = 1
1. 0.055
2. 0.270
3. y - 0.707 = -0.177(x - 1)
estimating the derivative from tables:
the function must be _______________ to estimate a derivative. this just means, the graph is ___________ and _________.
1. x (hours): 0, 2, 4, 7, 11; f(x) (miles): -2, 3, 10, 1, -3; using the table, estimate f'(3)
2.
differentiable
continuous
smooth
1. 7/2 miles/hr
2. -5 gallons/second^2
estimate the derivative at the given point by using a calculator.
1. f(x) = x?(2 - x); find f'(-10)
2. f(x) = sec(5x); find f'(2)
3. f(x) = ln(?x); find f'(1)
4. f(x) = e^(x/3); find f'(4)
5. f(x) = tan(sinx); find f'(1.3)
6. f(x) = 2^(ln(x)); find f'(2)
1. 4.907
2. -3.864
3. 0.500
4. 1.265
5. -1.010
6. 0.560
7. 8.098 m/sec
8. -0.132 ft/min
9. 0.153 stock/day
for each function, write the equation of the tangent line at the given value of x.
10. f(x) = ln2x / 4x at x = 1
11. f(x) = cos(tan(x)) at x = 2
12. f(x) = x^4 / ?x at x = 3
13. f(x) = x^2*sin(1/x) at x = 7
10. y - 0.173 = 0.077(x - 1)
11. y + 0.576 = 4.719(x - 2)
12. y - 46.765 = 54.560(x - 3)
13. y - 6.976 = 1.003 (x - 7)
use the tables to estimate the value of the derivative at the given point. indicate units of measure.
14 through 19 a and b
look at the practice in 2.3 estimating derivatives
differentiability
the derivative exists for each point in the domain. the graph must be a smooth line or curve for the derivative to exist. in other words, the graph looks like a line if you zoom in.
zoom in = looks like linear line
the derivative fails to exists where the function has a
discontinuity (hole, jump, or vertical asymptote)
corner or cusp (where the line squishes into basically one line and dead flowers for lines as they extend from the squish)
vertical tangent (part of the line is completely vertical, so its tangent line is
identify points where the function below is not continuous and/or not differentiable
look on page 1 of the 2.4 differentiability and continuity notes
true or false:
differentiability implies continuity
continuity implies differentiability
true
false
identify any x-values of the function that are not continuous and/or not differentiable
1 through 3
look at the practice page of 2.4 differentiability and continuity
derivative
equation for slope of tangent line (slope of function at exact point)
function:
f(x) = x^2
f(x) = x^3
f(x) = x^4
f(x) = x^5
find the function's derivative:
function's derivative:
f'(x) = 2x
f'(x) = 3x^2
f'(x) = 4x^3
f'(x) = 5x^4
the power rule:
f(x) = x^n
f'(x) = nx^(n-1)
find derivative:
1. y = x^37
2. y = x^9
y' = 37x^36
y' = 9x^8
rewrite, differentiate, and simplify
3. function: y = 1/x
4. function: y = 1/x^4
5. function: y = ?x
6. function y = ^7?x^3
3. rewrite: y = x^-1
differentiate: dy/dx = -1x^-2
simplify: dy/dx = -1/x^2
4. rewrite: y = x^-4
differentiate: y' = -4x^-5
simplify: y' = -4/x^5
5. rewrite: y = x^(1/2)
differentiate: y' = 1/2x^(-1/2)
simplify: y' = 1/(2?x)
6. rewrite: y = x^(3/7)
differ
find derivative:
7. f(x) = x/?x; find f'(7)
8. f(x) = ^3?(x)*(x^3). find f'(8)
7. f'(7) = 1/2?7
^ this is the slope of the tangent line at x = 7
8. f'(8) = 1280/3
parallel tangent lines:
9. let f(x) = x^4 and g(x) = x^3. at what value(s) of x do the graphs of f and g have parallel tangent lines
f'(x) = g'(x)
4x^3 = 3x^2
4x^3 - 3x^2 = 0
x^2(4x-3) = 0
x^2 = 0
4x-3 = 0
x = 0 and x = 3/4
parallel =
perpendicular =
same slope
opposite reciprocal slope
find dy/dx:
1. y = x^7
2. y = x
3. y = x^?
4. y = 1/x^5
5. y = 1/^4?x
6. y = ^9?x^4
7. y = 3^?x
8. y = x^e
9. y = x/^3?x
10. y = x^2(^6?x^5)
1. y' = 7x^6
2. y' = 1
3. y' = ?x(?-1)
4. y' = -5/x^6
5. y' = -1/(4*(^4?x^5))
6. y' = 4/(9*(^9?x^5))
7. y' = 1/(3*(^3?x^2))
8. y' = ex^(e-1)
9. y' = 2/(3*(^3?x))
10. y' = 17/(6*(^6?x^11))
find f'(a) for each function at the given value of a.
11. f(x) = x^4; find f'(-1)
12. f(x) = ?x; find f'(16)
13. f(x) = 1/x^4; find f'(2)
14. f(x) = 1/(^3?x); find f'(27)
11. -4
12. 1/8
13. -1/8
14. -1/243
find the equation of the tangent line of each function at the given value of x.
15. y = x^3 at x = -2
16. f(x) = ^4?x^3 at x = 1
17. f(x) = 1/x^4 at x = 2
15. y - 8 = 12(x + 2)
16. y - 1 = 3/4(x - 1)
17. y - (1/16) = -1/8(x - 2)
when do the two functions listed have parallel tangent lines?
18. f(x) = x^2 and g(x) = x^5
19. f(x) = ?x and g(x) = x^3
18. x = 0 and x = ^3?2/5
19. x = 0.488
derivative rules:
constant: (d/dx)*c =
constant multiple: (d/dx)*cu =
sum/difference (d/dx)(u+/-v) =
0
ex: y = e^2; y' = 0
c*(du/dx)
ex: y = (x^5)/3, y = 1/3x^5, y' = 5/3x^4; or y = 2x^3, y' = 3*2x = 6x
du/dx +/- du/dx
find the derivative of each function:
1. y = 2x^2 - 5/x + 6
2. y = 8?x - (x^6)/3 + 2pi^5
1. y' = 4x - 5/x^2
2. y' = 4/?x - 2x^5
horizontal tangent lines:
when does a function have a horizontal tangent line?
the slope of a horizontal tangent line is zero. to find where a function has a horizontal tangent line, we set the derivative equal to zero.
generally are min/max
3. find the x-values of any horizontal tangent lines of f(x) = 4x^2 + 7x - 13
f'(x) = 8x + 7
x = -7/8
normal lines:
a normal lines...
goes through the same point the tangent line does, but it is perpendicular to the tangent line
find the equation of a normal line of f(x) = x^3 - 4x^2 + x + 3 at x = 3
f(3) = -3
f'(x) = 3x^2 - 8x + 1
f'(3) = 4
y + 3 = -1/4(x - 3)
is the function f(x) = {5x^2 + 3x + 2, x<-1; -7x - 3, x>=-1 differentiable at x = -1?
yes
if it is not continuous (the two equations do not equal the same number when set equal to each other), then it is not...
differentiable (set derivatives equal to eachother)
what values of a and b would make this function f(x) = [x^2 - ax + 2, x<3; x+b, x>=3 differentiable at x = 3?
a = 5
b = -7
solve for b by setting equations equal to each other
b will still be an equation
then set derivatives equal to each other and solve for a
plug at into b to get real b
find the derivative of each function.
1. f(x) = 2x^3 - 4x + 5
2. g(x) = 5x^-2 - 1/2x^4
3. y = 2e^4 - 3x
4. y = ?x^2 - ?
5. y = 3x^2 - 1/6x^2
6. h(x) = (x^6)/3 + 6x^(2/3) - 4x^(1/2) + 2
7. f(x) = 1/?x + 3/5x
8. f(x) = ?x + 3*(^3?x) + 2
9. f(x) = 3x^7 - 4x^
1. f'(x) = 6x^2 - 4
2. g'(x) = -10/x - 2x^3
3. y' = -3
4. y' = 2?x
5. y' = 6x(1/(3x^3))
6. h'(x) = 2x^5 + 4/^3?x - 2/?x
7. f'(x) = -1/2?x^3 + -3/5x^2
8. f'(x) = 1/2?x + 1/(^3?x^2)
9. f'(x) = 21x^6 - 12x^2 + 5
10. y' = 2/?x
find the x-value(s) where the function has a horizontal tangent
11. f(x) = (x^3)/3 + 4x^2 + 12x - 13
12. f(x) = (x^4)/2 + x^3 + (x^2)/2 + 7
13. f(x) = (x^4)/4 - (10x^3)/3 + 21/x + 6/5
11. x = -6, x= -2
12. x = 0, x = -1/2, x = -1
13. x = 0, x = 7, x = 3
find the equations of the tangent and normal lines of each function at the given value of x.
14. f(x) = 3?x + 4 at x = 4
15. y = (x^2)/2 +3/2x - 2 at x = 8
16. f(x) = -x^3 + 2x^2 - 2 at x = 2
14. tangent: y - 10 = 3/4(x - 4); normal: y - 10 = -4/3(x - 4)
15. tangent: y - 42 = 19/2(x - 8); normal: y - 42 = -2/19(x - 8)
16. tangent: y + 2 = -4(x - 2); normal: y + 2 = 1/4(x - 2)
are the functions differentiable at the given value of x?
17. at x = 5. f(x) = {2x - 8/5x^2 + 10, x<=5; 50 - 14x, x>5
18. at x = 9. f(x) = {30/?x - x^2, x<9; x^2 - 5x - 107, x>=9
19. at x = 3. f(x) = {5x^2 - 2x + 1, x<=3; 3x^2 + 2x + 6, x>3
17. yes
18. no
19. no
what values of a and b would make the function differentiable at the given value of x?
21. at x = 2. f(x) = {ax^4 + x + 4, x<2; bx - 5, x>=2
22. at x = 1. f(x) = {a/x^2 + x^3 - 2, x<=1; x^2 + bx + 1, x>1
21. a = 3/16, b = 7
22. a = 4/3, b = -5/3
derivatives of cosx and sinx
d/dx cosx = -sinx
d/dx sinx = cosx
f'(x) if f(x) = 2sinx - 5cosx
f'(x) = 2cosx + 5sinx
ln1=
ln0=
e^0=
e^(lna)=
lne^a=
0
undefined
1
a
a
derivatives of exponential functions:
d/dx a^x =
d/dx e^x =
a^(x)lna
e^(x)lne = e^x
find f'(x) if f(x) = 2^x + 3e^x
f'(x) = 2^(x)ln2 + 3e^x
derivatives of logarithmic functions
d/dx log a x =
d/dx lnx = d/dx log e x =
1/x * 1/lna
1/x * 1/lne = 1/x
find f'(x) if f(x) = log 4 x - 4lnx
f'(x) = 1/xln4 - 4/x
find the derivative of each function:
1. f(x) = 2sinx + 5e^x
2. f(x) = 3^x - 4cosx
3. f(x) = log 2 x - sinx
1. f'(x) = 2cosx + 5e^x
2. f'(x) = 3^(x)ln3 + 4sinx
3. f'(x) = 1/(xln2) - cosx
Find the value of the derivative at the given point.
4. if f(x) = 3lnx + e^x. find f'(5)
f'(5) = 3/5 + e^5
if f(x) = 3cosx + sinx / 2, find f'(?)
f'(?) = -1/2
find the derivative of each function:
1. f(x) = 2lnx
2. f(x) = 5^x
3. f(x) = 9cosx
4. f(x) = 5e^x
5. f(x) = 8lnx - 4cosx + e
6. f(x) = 15sinx = 3e^x
7. f(x) = log 2 x
8. f(x) = log 7 x + cosx
9. f(x) = 3^x + 3x + x^3
10. f(x) = 3sinx
1. f'(x) = 2/x
2. f'(x) = 5^(x)ln5
3. f'(x) = -9sinx
4. f'(x) = 5e^x
5. f'(x) = 8/x + 4sinx
6. f'(x) = 15cosx - 3e^x
7. f'(x) = 1/xln2
8. f'(x) = 1/xln7 - sinx
9. f'(x) = 3^(x)ln3 + 3 + 3x^2
10. f'(x) = 3cosx
find the value of the derivative at the given point
11. if f(x) = 3x - 6cosx, find f'(?/2)
12. if f(x) = ?(x) - 2lnx, find f'(4)
13. if f(x) = 4e^x + 5sinx, find f'(0)
14. if f(x) = 2cosx + e^x, find f'(?)
11. 9
12. -1/4
13. 9
14. e^?
find the equation of the tangent line at the given x-value
15. f(x) = 3cosx + x at x = ?/2
16. f(x) = 4e^x - 3sinx + x^2 at x = 0
15. y - ?/2 = -2(x - ?/2)
16. y - 4 = x
product rule
h(x) = f*g
h'(x) = d/dx[f(x)g(x)] = (f(x)
d/dx[g(x)]) + (g(x)
d/dx[f(x)])
*you don't need the product rule if it doesn't have an "x
find the derivative of each function:
1. f(x) = 8xsinx
2. g(x) = 2e^(x)*(?x)
3. h(x) = ((1/x) + 1)(2x^2 - 5)
1. f'(x) = 8sinx+8xcosx
2. g'(x) = e^(x)*((2x+1)/?x)
3. h'(x) = 5/(x^2) + 4x + 2
the table below shows values of two differentiable functions f and g, as well as their derivatives.
x: 2, -5
f(x): 4, 3
f'(x): -2, 4
g(x): -1, -2
g'(x): 2, 5
4. h(x) = 3f(x)g(x); find h'(2)
5. r(x) = ((f(x)/2) + 2)(3-g(x)); find r'(-5)
4. 30
5. -7.5
find the derivative of each function:
1. f(x) = (2x-3)cosx
2. g(x) = 2x^(3)e^x
3. h(x) = 4?(x)lnx
4. f(x) = (4-5x)cosx
5. g(x) = 6lnxsinx
6. h(x) = 2e^x(x^2 + x)
7. f(x) = 8sinxcosx
8. g(x) = (3/x)lnx
9. h(x) = 2x^(5)cosx
10. f(x) = e^(x)sinx
1. f'(x) = 2sinx + (2x - 3)cosx
2. g'(x) = 6x^(2)e^x + 2x^(3)e^x
3. h'(x) = (2lnx/?x) + (4?x/x)
4. f'(x) = 5cosx - (4 - 5x)sinx
5. g'(x) = (6/x)(sinx) + 6lnxcosx
6. h'(x) = 2e^(x)(x^2 + 3x + 1)
7. f'(x) = 8cos^2x - 8sin^2x
8. g'(x) = (-3lnx + 3)/x^2
9. h'
use the table to find the value of the derivatives of each function.
11. x:7, f(x):-5, f'(x): 3, g(x): 2, g'(x): -3.
a. h(t) = f(x)g(x); find h'(7)
b. m(x) = 5f(x)g(x); find m'(7)
c. s(x) = (3f(x) - 1)(g(x) + 2); find s'(7)
12. x:-4, a(t):2, a'(t):-3, b(t
11.
a. 21
b. 105
c. 84
12.
a. 14
b. -42
c. -33
13.
a. -17
b. 17
c.-17/2
find the equation of the tangent line at the given x-value
14. f(x) = 8sinxcosx at x = ?/3
15. g(x) = -2xe^x at x = 0
14. y - 2?3 = -4(x - ?/3)
15. y = -2x
quotient rule
h(x) = f/g
h'(x) = d/dx [f(x)/g(x)] = ((g(x)
d/dx[f(x)]) - (f(x)
d/dx[g(x)])) / g(x)^2
find the derivative of each function:
1. y = 2x^2 / 3x+1
2. g(x) = 3e^x / 2x
3. h(x) = sinx / 2x^(2)-5
4. h(x) = 3x+1 / 2x^2
1. y' = 6x^(2)+4x / (3x+1)^2
2. g'(x) = 6e^(x)(x-1) / 4x^2
3. h'(x) = (cosx2x^(2)-5 - 4xsinx) / (2x^(2) - 5)^2
4. h'(x) = (-3/2x^2) - (1/x^3)
the table below shows values of two differentiable functions f and g, as well as their derivatives
x: 2
f(x): 4
f'(x): -2
g(x): -1
g'(x) = 2
5. h(x) = f(x)/3g(x); find h'(2)
6. r(x) = -(g(x) / (1-f(x))); find r'(2)
5. -2
6. 1/3
find the derivative of each function:
1. h(x) = 4x-1 / 3x+2
2. g(x) = sinx / x
3. h(x) = (x^3 = 3x^2 - x) / 2x
4. h(x) = 4x/lnx
5. (3x^4 - 2x^2 - 3?x) / x
6. g(x) = 2x^5 / cosx
7. e^x / 4sinx
8. 2x+4 / 3x+2
9. (x^3 + 3x^2 - x) / x^2
1. h'(x) = 11 / (3x+2)^2
2. g'(x) = (xcosx - sinx) / x^2
3. h'(x) = x + 1
4. h'(x) = (4lnx - 4)/(lnx)^2
5. f'(x) = (3x^4 - 2x^2 - 3?x) / x
6. g'(x) = 2x^5 / cosx
7. f'(x) = e^x / 4sinx
8. f'(x) = -8x / (3x + 2)^2
9. g'(x) = 1 + 1/x^2
use the table to find the value of the derivatives of each function
10. x:7, f(x):-5, f'(x):3, g(x):2, g'(x):-3
a. h(t) = 5f(x) / g(x); find h'(7)
b. m(x) = g(x)+2 / 3f(x)
11. t:-4, a(t):2, a'(t):-3, b(t):-4, b'(t):1
a. f(t) = -b(t) / 3a(t): find f'(-4)
b
10.
a. -45/4
b. 1/25
11.
a. 5/6
b. -13/25
12.
a. 1/8
b. 7/2
find the equation of the tangent line at the given x-value
13. f(x) = sinx/cosx at x = ?/3
14. g(x) = -2x/e^x at x = 0
13. y - ?3 = 4(x - (?/3)
14. y = -2x
d/dxsinx =
cosx
d/dxcosx =
-sinx
d/dxtanx =
sec^2x
d/dxcotx =
-csc^2x
d/dxsecx =
secxtanx
d/dxcscx =
-cscxcotx
if it starts with c...
it is negative
find the derivative of y = sinxtanx
y' = sinx + sinxsec^2x
find f(?/6) if f(x) = x / secx
(6?3 - ?) / 12
estimate the derivative with a calculator of g(x) = csc^(2)4x at x = 2
1.202
csc =
1/sin
sec =
1/cos
find the derivative of each function (some of which are at a given x-value)
1. y = 5 - cscx
2. h(x) = 2xtanx
3. r = sin?/?
4. g(x) = cotx/x
5. f(x) = 1/2cosx
6. y = 5xsecx
7. f(x) = 3tanx at x = 2?/3
8. f(x) = 2secx at x = ?/4
9. f(x) = xcotx at x = ?/6
1. y' = cscxcotx
2. h'(x) = 2tanx + 2xsec^2x
3. r' = (cos?(?) - sin?) / ?^2
4. g'(x) = (-xcsc^2x - cotx) / x^2
5. f'(x) = sinx/2cos^2x
6. y' = 5secx + 5xsecxtanx
7. 12
8. 2?2
9. ?3 - 2?/3
estimate the derivative at the given x-value by using a calculator
10. f(x) = sin^(2)(x/5) at x = 1.8
11. f(x) = cot(x^2) / 2 at x = -1
12. f(x) = 3sec(e^x) at x = 2.5
10. 0.132
11. 1.412
12. -15.923
find the equations of both the normal line and the tangent line
13. y = secx at x = ?
14. y= tanx at x = ?/3
13. tangent: y = -1
normal: ? possibly x = ? but prolly not
14. tangent: y - ?3 = 4(x - ?/3)
normal: y - ?3 = -1/4(x - ?/3)
find the average rate of change of each function on the given interval
1. w(x) = lnx; 1 <= x <= 7
2. s(t) = -t^2 - t + 4; [1,5]
3. find the derivative of y = 2x^2 + 3x - 1 by using the definition of the derivative.
4. for the function h(t), h is the tempe
1. 0.324
2. -7 ft/sec
3. y' = 4x + 3
4. a. After 15 minutes, the temperature of the oven is 420 degrees F
b. At the 43th minute, the oven temperature is decreasing by 11 degrees F/min
find the derivative of each function
5. f(x) = 4 - (1/(2x^2))
6. g(x) = 3?x - (6/x^2) + 5?^3
7. h(x) = 4e^x - 2cosx
8. s(t) = t^2sin(t)
9. d(t) = 3?(t)lnt
10. y = 4/x - secx
11. h(x) = (2-x)/(x+2)
5. f'(x) = 1/x^3
6. g'(x) = (3/2?x) + (12/x^3)
7. h'(x) = 4e^x + 2sinx
8. s'(t) = 2tsin(t) + (t^2)(cos(t))
9. d'(t) = (3lnt/2?x) + (3?t/t)
10. y' = -4/x^2 - secxtanx
11. h'(x) = -4/((x+2)^2)
find the equation of the tangent line of the function at the given x-value
12. f(x) = -2x^3 + 3x at x = -1
13. f(x) = 4sinx - 2 at x = ?
14. find the equation for the normal line of y = 1/2x^2 + 3/4x - 4 at x = -3
15. if f(x) = 3sinx - 2e^x find f'(0)
12. y + 1 = -3(x + 1)
13. y + 2 = -4(x - ?)
14. y + 7/9 = 4/9(x + 3)
15. 1
16. if f(x) = sin(3x^2 - 2); find f'(7)
17. if f(x) = csc(3x) at x = 2
18. use the table below to estimate the value of d'(120).
t seconds: 2, 13, 60, 180, 500
d(t) feet: 10, 81, 412, 808, 2105
19. is the function differentiable at x = 2?
f(x) = {3x - 3x^
16. 260.246
17. -36.899
18. 3.3ft/sec
19. yes
20. a = 47/11; b = -13/44