Index FAQ The derivative as the slope of the tangent line (at a point)
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Transcript of Index FAQ The derivative as the slope of the tangent line (at a point)
![Page 1: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/1.jpg)
Index FAQ
The derivative as the slope of the tangent line
(at a point)
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Index FAQ
Video help: MIT!!! http://ocw.mit.edu/courses/
mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/
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Index FAQ
What is a derivative?
A function, which gives the:the rate of change of a
function in generalthe slope of the line tangent
to the curve in general
![Page 4: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/4.jpg)
Index FAQ
What is a differential quotient?
Just a number!the rate of change of a function at a
given pointthe slope of the line tangent to the
curve at a certain pointThe substitutional value of the
derivative
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Index FAQ
The tangent line
single pointof intersection
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Index FAQ
slope of a secant line
ax
f(x)
f(a)
f(a) - f(x)
a - x
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Index FAQ
slope of a (closer) secant line
ax
f(x)
f(a)
f(a) - f(x)
a - x
x
![Page 8: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/8.jpg)
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closer and closer…
a
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Index FAQ
watch the slope...
![Page 10: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/10.jpg)
Index FAQ
watch what x does...
ax
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Index FAQ
The slope of the secant line gets closer and closer to the slope of the tangent line...
![Page 12: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/12.jpg)
Index FAQ
As the values of x get closer and closer to a!
ax
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Index FAQ
The slope of the secant lines gets closer
to the slope of the tangent line...
...as the values of x get closer to a
Translates to….
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limax
f(x) - f(a)x - a
Equation for the slope
Which gives us the the exact slope of the line tangent to the curve at a!
as x goes to a
Differential quotient
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Differential quotient: other form
aa+h
f(a+h)
f(a)
f(x+h) - f(x)
(x+h) - x= f(x+h) - f(x)
h
(For this particular curve, h is a negative value)
h
limh0
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Index FAQ
Rates of Change:
Average rate of change = f x h f x
h
Instantaneous rate of change = 0
limh
f x h f xf x
h
These definitions are true for any function.
Velocity and other Rates of Change
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Consider a graph of displacement (distance traveled) vs. time.
time (hours)
distance(miles)
Average velocity can be found by taking:change in position
change in time
s
t
t
sA
B
ave
f t t f tsV
t t
The speedometer in your car does not measure average velocity, but instantaneous velocity.
0
limt
f t t f tdsV t
dt t
(The velocity at one moment in time.)
Velocity and other Rates of Change- physical menaing of the differential quotient
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Velocity and other rates of change
Velocity is the first derivative of position.
Acceleration is the second derivative of position.
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Example:Free Fall Equation
21
2s g t
GravitationalConstants:
2
ft32
secg
2
m9.8
secg
2
cm980
secg
2132
2s t
216 s t 32 ds
V tdt
Speed is the absolute value of velocity.
Velocity
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Acceleration is the derivative of velocity.
dva
dt
2
2
d s
dt example:
32v t
32a If distance is in: feet
Velocity would be in:feet
sec
Acceleration would be in:
ftsec sec
2
ft
sec
3.4 Velocity and other Rates of Change
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Index FAQtime
distance
acc posvel pos &increasing
acc zerovel pos &constant
acc negvel pos &decreasing
velocityzero
acc negvel neg &decreasing acc zero
vel neg &constant
acc posvel neg &increasing
acc zero,velocity zero
Velocity and other Rates of Change
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Index FAQ
To be differentiable, a function must be continuous and smooth.Derivatives will fail to exist at:
corner
f x x
cusp
2
3f x x
vertical tangent
3f x x
discontinuity
1, 0
1, 0
xf x
x
Differentiability
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Theorem : f is differentiable on the interval (a,b). f is continuous on the interval (a,b).
Proof: Assume that f ’(c) exists for any c in (a,b).
Then lim [ f(c+h)- f(c)] . h0
= f ’(c) • 0 = 0
f ’(c) = lim f(c+h) -f(c) . h 0 h
= f ’(c) • lim h . h 0
So lim [ f(c+h) - f(c)] = 0 . h0
, and from here we get lim f(c+h) = f(c) . . h0
So f is continuous at c for every c in (a,b).
/ • h
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Example: Since the derivative of f(x)= 5x2+x+1 is f ’(x) = 10x+1, which exists for every real number x. So f(x)= 5x2+x+1 is continuous everywhere.
RemarkThe reverse of this theorem is not
true.
Counter example: We know that f(x) = |x| is continuous on R , but at x=0 it’s not differentiable since:
lim l0+hl –l0l h 0 h
= lim lhl . h 0 h
, which approaches to +1 if h 0 –1 if h0
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Index FAQ
To be differentiable, a function must be continuous and smooth.Derivatives will fail to exist at:
corner
f x x
cusp
2
3f x x
vertical tangent
3f x x
discontinuity
1, 0
1, 0
xf x
x
Differentiability
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Index FAQ
If the derivative of a function is its slope, then for a constant function, the derivative must be zero. There is no change...
0dc
dx
example: 3y
0y
The derivative of a constant is zero.
Derivatives of some elementary functions
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Index FAQ
We saw that if , .2y x 2y x
This is part of a pattern.
1n ndx nx
dx
examples:
4f x x
34f x x
8y x
78y x
power rule
Derivatives of some elementary functions
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Index FAQ
Find the horizontal tangents of:
4 22 2y x x 34 4
dyx x
dx
Horizontal tangents occur when slope = zero.
34 4 0x x 3 0x x
2 1 0x x
1 1 0x x x
0, 1, 1x
Substituting the x values into the original equation, we get:
2, 1, 1y y y
(The function is even, so we only get two horizontal tangents.)
Rules for Differentiation
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Index FAQ
Rates of Change:
Average rate of change = f x h f x
h
Instantaneous rate of change = 0
limh
f x h f xf x
h
These definitions are true for any function.
( x does not have to represent time. )
Velocity and other Rates of Change
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2
0
2
Consider the function siny
We could make a graph of the slope: slope
1
0
1
0
1Now we connect the dots!The resulting curve is a cosine curve.
sin cosd
x xdx
Derivatives of Trigonometric Functions
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Index FAQ
Derivatives of Trigonometric Functions
h
xsin)hxsin(lim)'x(sin
0h
h
xsinxcoshsinhcosxsinlim
0h
h
xcoshsinlim
h
)1h(cosxsinlim
0h0h
h
xcoshsin)1h(cosxsinlim
0h
Proof
h
xh
h
hxx
dx
dhh
cossinlim
)1(cossinlimsin
00
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Index FAQ
Derivative of the cosine Function
h
xcos)hxcos(lim)'x(cos
0h
h
xsinhsinlim
h
)1h(cosxcoslim
0h0h
h
xsinhsin)1h(cosxcoslim
0h
Find the derivative of cos x:
h
xsinhsinlim
h
)1h(cosxcoslim
0h0h
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Derivative of the cosine function is sine (cont.)
xsin1.xsin0.xcosh
hsinlimxsin
h
)1h(coslimosxc
h
xsinhsin
h
)1h(cosxcoslim
h
xsinhsin)1h(cosxcoslim
h
xcosxsinhsinhcosxcoslim
0h0h
0h
0h
0h
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We can find the derivative of tangent x by using the quotient rule.
tand
xdx
sin
cos
d x
dx x
2
cos cos sin sin
cos
x x x x
x
2 2
2
cos sin
cos
x x
x
2
1
cos x2sec x
2tan secd
x xdx
Derivatives of Trigonometric Functions
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Derivatives of the remaining trig functions can be determined the same way.
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
Derivatives of Trigonometric Functions
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Index FAQ
The Derivatives of the Sum, Difference, Product and Quotient
If and are derivable, and is any constant, u x v x C
then so is , , , and
. Its derivative is given by the formula
u x v x u x v x Cu x
u x
v x
(1) ( ) ( ) ( ) ( )u x v x u x v x
(2) ( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v x
(3) ( ) ( )Cu x Cu x 2
( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)
( ) ( )
u x u x v x u x v xv x
v x v x
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Index FAQ
Proof
(1) Let ( ) ( ), we have to examiney u x v x
0 0
( ) ( ) ( ) ( )lim limx x
y u x x v x x u x v x
x x
0
( ) ( ) ( ) ( )limx
u x x u x v x x v x
x
(1) ( ) ( ) ( ) ( )u x v x u x v x
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Index FAQ
0lim( ) ( ) ( )x
u vu x v x
x x
Thus ( ) ( ) is derivable and
( ) ( ) ( ) ( )
u x v x
u x v x u x v x
A similar argument applies to ( ) ( ),
that is
( ) ( ) ( ) ( )
u x v x
u x v x u x v x
(1) ( ) ( ) ( ) ( )u x v x u x v x
![Page 39: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/39.jpg)
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x 0
(2) Let ( ) ( ), then we express in terms
of and . Finally, we determine by
examining lim
y u x v x y
u v y x
y
x
0 0
( ) ( ) ( ) ( )lim limx x
y u x x v x x u x v xy
x x
0
[ ( ) ][ ( ) ] ( ) ( )limx
u x u v x v u x v x
x
(2) ( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v x
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Index FAQ
0
( ) ( )limx
u x v v x u u v
x
0lim[ ( ) ( ) ]x
u v vv x u x ux x x
( ) ( ) ( ) ( )u x v x u x v x
Thus, ( ) ( ) is derivable and
( ) ( ) ( ) ( ) ( ) ( )
u x v x
u x v x u x v x u x v x
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Index FAQ
HOMEWORK!!
(3) ( ) ( )Cu x Cu x
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2
( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)
( ) ( )
u x u x v x u x v xv x
v x v x
0 0
( ) ( )( ) ( )
lim limx x
u x x u xy v x x v x
yx x
0
( ) ( )( ) ( )
limx
u x u u xv x v v x
x
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0 0
( ) ( )( ) ( )
lim limx x
u x x u xy v x x v x
yx x
0
( ) ( )( ) ( )
limx
u x u u xv x v v x
x
2
( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)
( ) ( )
u x u x v x u x v xv x
v x v x
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0
[ ( ) ] ( ) ( )[ ( ) ]lim
[ ( ) ] ( )x
u x u v x u x v x v
v x v v x x
0
( ) ( )lim
[ ( ) ] ( )x
uv x u x v
v x v v x x
0
( ) ( )lim
[ ( ) ] ( )x
u vv x u xx xv x v v x
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dy dy du
dx du dx
Chain Rule:
example: sinf x x 2 4g x x Find: at 2f g x
cosf x x 2g x x 2 4 4 0g
0 2f g cos 0 2 2 1 4 4
Chain Rule
If is the composite of and , then:f g y f u u g x
at at xu g xf g f g )('))((' xgxgf
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Remark
f(g(x))’= f ’(g(x)) g’(x) says that to get the
derivative of the “nested functions” you multiply
the derivative of each one starting from left to
right and so on
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Example : Find y’(1) for y = (3x2-2)3( 5x3-x-3)4
y ’= 3(3x2 -2)2 (3x2-2)’( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 ( 5x3-x-3)’
y ’= 3(3x2 -2)2 (6x) ( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 (15x2-1)
y ’(1) = 3(3-2)2 (6) (5-1-3)4 + (3-2)3 4 (5-1-3)3 (15-1)
YOUR TURN!, find when x=1 .
= 74
dy .
dx
. 2 x - 1 √5x2+4
For y =
Example for using Chain rule
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2sin 4y x
2 2cos 4 4d
y x xdx
2cos 4 2y x x
Example for using Chain rule
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2cos 3d
xdx
2cos 3
dx
dx
2 cos 3 cos 3d
x xdx
2cos 3 sin 3 3d
x x xdx
2cos 3 sin 3 3x x
6cos 3 sin 3x x
The chain rule can be used more than once.
(That’s what makes the “chain” in the “chain rule”!)
Example for using Chain rule
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2 2 1x y This is not a function, but it would still be nice to be able to find the slope.
2 2 1d d dx y
dx dx dx Do the same thing to both sides.
2 2 0dy
x ydx
Note use of chain rule.
2 2dyy xdx
2
2
dy x
dx y
dy x
dx y
Implicit Differentiation
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Index FAQ
22 siny x y 22 sin
d d dy x y
dx dx dx
This can’t be solved for y.
2 2 cosdy dy
x ydx dx
2 cos 2dy dy
y xdx dx
22 cosdy
xydx
2
2 cos
dy x
dx y
This technique is called implicit differentiation.
1 Differentiate both sides w.r.t. x.2 Solve for y’
Implicit Differentiation
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Implicit Differentiation
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect to x.
2. Collect the terms with y’=dy/dx on one side of the equation.
3. Factor out y’=dy/dx .
4. Solve for y’=dy/dx .
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Index FAQ
Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
2 2 7x xy y
2 2 0dydy
x yx ydxdx
Note product rule.
2 2 0dy dy
x x y ydx dx
22dy
y xy xdx
2
2
dy y x
dx y x
2 2 1
2 2 1m
2 2
4 1
4
5
Implicit Differentiation
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Index FAQ
Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
4
5m tangent:
42 1
5y x
4 42
5 5y x
4 14
5 5y x
normal:
52 1
4y x
5 52
4 4y x
5 3
4 4y x
Implicit Differentiation
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Index FAQ
Find if .2
2
d y
dx3 22 3 7x y
3 22 3 7x y 26 6 0x y y
26 6y y x 26
6
xy
y
2x
yy
2
2
2y x x yy
y
2
2
2x xy y
y y
2 2
2
2x xy
y
x
yy
4
3
2x xy
y y
Substitute back into the equation.
y
Implicit Differentiation
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Index FAQ
siny x
1siny xWe can use implicit differentiation to find:
1sind
xdx
1siny x
sin y xsin
d dy x
dx dx
cos 1dyydx
1
cos
dy
dx y
Derivatives of Inverse Trigonometric Functions
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Index FAQ
We can use implicit differentiation to find:
1sind
xdx
1siny x
sin y xsin
d dy x
dx dx
cos 1dyydx
1
cos
dy
dx y
2 2sin cos 1y y
2 2cos 1 siny y 2cos 1 siny y
But2 2
y
so is positive.cos y
2cos 1 siny y
2
1
1 sin
dy
dx y
2
1
1
dy
dx x
Derivatives of Inverse Trigonometric Functions
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Index FAQ
1siny x
1
cos
dy
dx y
Derivatives of Inverse Trigonometric Functions
)cos(sin
11 xdx
dy
21
1
xdx
dy
xy sin
1cos dx
dyy
![Page 59: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/59.jpg)
Index FAQ
Derivatives of Inverse Trigonometric Functions
)(tansec
112 xdx
dy
21
1
xdx
dy
xy tan
1sec2 dx
dyy
Find xdx
d 1tan
xy 1tan
ydx
dy2sec
1
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Index FAQ
Look at the graph of xy e
The slope at x = 0 appears to be 1.
If we assume this to be true, then:
0 0
0lim 1
h
h
e e
h
definition of derivative
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
Now we attempt to find a general formula for the derivative of using the definition.
xy e
0
limx h x
x
h
d e ee
dx h
0lim
x h x
h
e e e
h
0
1lim
hx
h
ee
h
0
1lim
hx
h
ee
h
1xe xe
This is the slope at x = 0, which we have assumed to be 1.
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
xe is its own derivative!
If we incorporate the chain rule: u ud due e
dx dx
We can now use this formula to find the derivative ofxa
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
xda
dx
ln xade
dx
lnx ade
dx ln lnx a d
e x adx
Incorporating the chain rule:
lnu ud dua a a
dx dx
Derivatives of Exponential and Logarithmic Functions
aaadx
d xx ln
![Page 64: Index FAQ The derivative as the slope of the tangent line (at a point)](https://reader033.fdocuments.us/reader033/viewer/2022052510/56649de85503460f94ae2185/html5/thumbnails/64.jpg)
Index FAQ
So far today we have:
u ud due e
dx dx lnu ud du
a a adx dx
Now it is relatively easy to find the derivative of .ln x
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
lny xye x
yd de x
dx dx
1y dyedx
1y
dy
dx e
1ln
dx
dx x
1ln
d duu
dx u dx
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
To find the derivative of a common log function, you could just use the change of base rule for logs:
logd
xdx
ln
ln10
d x
dx
1ln
ln10
dx
dx
1 1
ln10 x
The formula for the derivative of a log of any base other than e is:
1log
lna
d duu
dx u a dx
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
u ud due e
dx dx lnu ud du
a a adx dx
1log
lna
d duu
dx u a dx
1ln
d duu
dx u dx
Derivatives of Exponential and Logarithmic Functions
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Index FAQ
Derivatives of Exponential and Logarithmic Functions
Logarithmic differentiation
Used when the variable is in the base and the exponent
y = xx
ln y = ln xx
ln y = x ln x
xx
xdx
dy
yln
11
xydx
dyln1
xxdx
dy x ln1