Unformatted text preview: MTH405 Wk 8 Lect 1 Product Rule The Product Rule deals with functions multiplied together, i.e.,
y = f (x) Â· g(x).
Theorem. For multiplication of two functions have y = f (x) Â· g(x), we d
[f (x) Â· g(x)] = [f 0 (x) Â· g(x)] + [g 0 (x) Â· f (x)]
dx or in a more usable form y 0 = f 0 g + g 0 f. Quotient Rule The Quotient Rule deals with functions divided with each other,
i.e.,
y= f (x)
g(x) where g(x) 6= 0.
Theorem. For division of two functions f (x)
, we have
g(x)
d f (x)
f 0 (x)g(x) âˆ’ g 0 (x)f (x)
=
dx g(x)
(g(x))2
or in a more usable y0 = f 0g âˆ’ g0f
.
g2 1 Chain Rule The Chain Rule deals with functions composed inside functions, i.e.,
y = f (g(x)).
Theorem. For composition of two functions y = f (g(x)), we denote u = g(x),
then we have y = f (u),
hence the Chain Rule is dy
dy du
=
Â· .
dx
du dx
Example. Dierentiate these functions. 1. y = x sin x = (x)(sin x) 2. f (x) = x2 ex = (x2 ) (ex ) 3. f (x) = (x2 + 3x)(x3 + 30x2 + 7x âˆ’ 4). 4. y= x sin x
ex 5. y= 2x+3
5x+10 6. y= ex
x2 7. y = sin x2 (u = x2 is the inside function and y = sin u is outside , function), 8. y = (x2 âˆ’ 40)20 (u = x2 âˆ’ 40 is the inside function and y = u20 9. y = e5x (u = 5x is the inside function and y = eu is the outside is outside function ),
function), 2 Implicit Dierentiation
Till now, we have been dierentiating functions which look nice and
organised, i.e., we have y on the left equals to a function on the right
expressed in terms of x. So we notice the function and choose a rule
to use. But what about this function?
y Â· tan3 (xy 2 + y) = x Ofcourse we can not use any of the rules since y is also a part of
the function. So our question is: How do we dierentiate functions
which are very complicated and x and y variables are jumbled up?
We use the technique of implicit dierentiation.
â€¢ â€¢ Explicit dierentiation is where in the function, you can separate the variables x and y . And y is alone on the left of
the equal sign and the rest of the function in terms of x is on
the right of the equal sign. Till now, we have covered all the
aspects of explicit dierentiation.
x2 +x
Example, y = sin x, y = 3x
2 +x and so on. In explicit dierentiation, we only dierentiate the function in terms of x. Implicit dierentiation is where in the function, you can not separate the variables x and y (or by separating them, the
function becomes too complicated to use). Then both x and
y are to be dierentiated. Note that whenever any y term is
dy
dierentiated, then multiply it with dx
. Example. Use the technique of implicit dierentiation to nd 1. 5y 2 + sin y = x2 . 2. xy = 1. 3. x3 + y 3 = 3xy . 4. xy 3
1+sec y = 1 + y4. 3 dy
dx . Applications of Dierentiation
Rates of Change  Distance, Velocity and Acceleration
Given a function which gives the distance travelled by an object
moving in a direction is denoted by
y = s(t) = d(t) = x(t). Note that the distance function is a function depending on t time.
Then the velocity function is given by
v(t) = and the acceleration dy
= s0 (t) = d0 (t) = x0 (t)
dt function is given by
d2 y
d dy
= 2 = s00 (t) = d00 (t) = x00 (t).
a(t) = v (t) =
dt dt
dt
0 The velocity function v(t) = x0 (t) gives the instantanous velocity of
the object at time t and the acceleration function a(t) = v 0 (t) gives
the instantaneous acceleration of the object at time t.
If y = x(t), then we dene the average velocity (average rate of
change of y with respect to t) over the interval [t0 , t1 ] to be
v ave = x(t1 ) âˆ’ x(t0 )
.
t1 âˆ’ t0 4 Supplies are dropped from a helicopter and the distance
fallen in a time t seconds is given by:
Example. 1
x(t) = gt2 ,
2 where g = 9.8m/s2
1. Determine the velocity function of the supplies. 2. Determine the acceleration function of the supplies. 3. 4. 5. 6. Determine the velocity and acceleration of the supplies after it
has fallen for 2 seconds?
After how many seconds does it hit the ground if the supplies
are dropped from 100m?
What is the velocity of the supplies as it hits the ground?
What is the average velocity of the supplies?
âˆ’xi
over the interval [ti , tf ]).
(Hint: vave = xtff âˆ’t
i Solution.
1. Direntiating x(t) once, we have
1 2
gt
2
v(t) = x0 (t) = gt
x(t) = 2. Dierentiating v(t) once, we have
v(t) = gt
a(t) = g = 9.8m/s2 . 3. Finding v(2) and a(2), we have
v(2) = g(2) = (9.8)(2) = 19.6m/s
a(2) = g = 9.8m/s2 . 5 4. Using the distance function to nd time of free fall, we have
1 2
gt
2
1
100 =
(9.8)t2
2
200
2
t =
9.8
t = 4.516 sec. x(t) = 5. Since it takes t = 4.516sec to hit the ground, we input it in the
velocity function
v(t) = gt
v(4.516) = (9.8)(4.516)
= 44.257m/s. 6. Average velocity
x(tf ) âˆ’ x(ti )
tf âˆ’ ti
100 âˆ’ 0
=
(4.516) âˆ’ 0
= 22.14m/s. vave = 6 Optimization (Maximization and Minimization)
Note that Optimization involves nding turning points to maximize
or to minimize.
A rectangular area is formed having a perimeter of 40cm.
Determine the length and width of the rectangle such that it encloses
the maximum area. Example. Solution. This is a maximization question since we are maximizing area. So let see what we know about this rectangle (Note
that in this question we can use all 4 sides of the rectangle).
Perimeter P = 40 = 2L + 2W
Area A = L Â· W
Our aim is to nd the values of L and W such that area is maximised.
So we use perimeter equation, and make W the subject. So
40 = 2L + 2W
W = 20 âˆ’ L and substitute into area equation
A = LÂ·W
= L(20 âˆ’ L)
A = 20L âˆ’ L2 Then dierentiating and then equating to 0, we have
dA
= 20 âˆ’ 2L
dL
0 = 20 âˆ’ 2L
L = 10cm. Then W = 10cm. 7 Determine the area of the largest piece of rectangular
ground that can be enclosed by 100m of fencing, if part of an existing
straight wall is used as one side.
Example. Solution. Note that in this question we can use 3 sides of the
rectangle since 1 side is backed by the wall so it does not require
fencing. So let see what we know about this rectangle.
Perimeter P = 100 = L + 2W
Area A = L Â· W
Our aim is to nd the values of L and W such that area is maximised.
So we use perimeter equation, and make W the subject. So
100 = L + 2W
1
W = 50 âˆ’ L
2 and substitute into area equation
A = LÂ·W
1
= L(50 âˆ’ L)
2
1 2
A = 50L âˆ’ L
2 Then dierentiating and then equating to 0, we have
dA
= 50 âˆ’ L
dL
0 = 50 âˆ’ L
L = 50cm. Then W = 50 âˆ’ 21 L = 50 âˆ’ 21 (50) = 25m. 8 ...
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 Spring '16
 Alveen Chand
 Derivative, Implicit Di1Berentiation

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