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THE DIFFERENTIALS
Consider a function defined by y=f(x) where x is
the independent variable. In the four-step rule we
introduced the symbol Δx to the denote the
increment of x. Now we introduce the symbol dx
which we call the differential of x. Similarly, we shall
call the symbol dy as the differential of y. To give
separate meanings to dx and dy, we shall adopt the
following definitions of a function defined by the
equation y=f(x).
DEFINITION 1: dx = Δx
In words, the differential of the independent
variable is equal to the increment of the variable.
DEFINITION 2: dy = f’ (x) dx
In words, the differential of a function is equal to
its derivative multiplied by the differential of its
independent variable.
We emphasize that the differential dx is also an
independent variable, it may be assigned any value
whatsoever. Therefore, from DEFINITION 2, we see
that the differential dy is a function of two
independent variables x and dx. It should also be
noted that while dx=Δx, dy≠Δy in general.
Suppose dx≠0 and we divide both sides of the
equation
dy = f’ (x) dx
by dx. Then we get
( )x'f
dx
dy
=
Note that this time dy/dx denotes the quotient of
two differentials, dy and dx . Thus the definition of
the differential makes it possible to define the
derivative of the function as the ratio of two
differentials. That is,
( )
xofaldifferentithe
yofaldifferentithe
dx
dy
x'f ==
The differential may be given a geometric
interpretation. Consider again the equation y=f(x)
and let its graph be as shown below. Let P(x,y) and
Q(x+Δx,f(x)+Δx) be two points on the curve. Draw the
tangent to the curve at P. Through Q, draw a
perpendicular to the x-axis and intersecting the
tangent at T. Then draw a line through P, parallel to
the x-axis and intersecting the perpendicular through
Q at R. Let θ be the inclination of the tangent PT.
P
Q
T
R
θ
From Analytic Geometry, we know that
slope of PT = tan θ
But triangle PRT, we see that
x
RT
PR
RT
tan
∆
θ ==
However, Δx=dx by DEFINITION 1 . Hence
dx
RT
tan =θ
But the derivative of y=f(x) at point P is equal to the
slope of the tangent line at that same point P.
slope of PT = f’(x)
Hence,
( )
dx
RT
x'f =
And , RT = f’(x) dx
But, dy = f’ (x) dx
Hence, RT = dy
We see that dy is the increment of the ordinate of
the tangent line corresponding to an increment in Δx
in x whereas Δy is the corresponding increment of
the curve for the same increment in x. We also note
that the derivative dy/dx or f’(x) gives the slope of
the tangent while the differential dy gives the rise of
the tangent line.
DIFFERENTIAL FORMULAS
Since we have already considered dy/dx as the
ratio of two differentials, then the differentiation
formulas may now be expressed in terms of
differentials by multiplying both sides of the
equation by dx. Thus
d(c) = 0
d(x) =dx
d(cu) = cdu
d(u + v) = du + dv
d(uv) = udv + vdu
d(u/v) = (vdu – udv)/v2
d(un
) = nun-1
du
( ) u2duud =
EXAMPLE 1: Find dy for y = x3
+ 5 x −1.
( )
( )dx53xdy
dx5dxx3
1x5xddy
2
2
3
+=
+=
−+=
EXAMPLE 2: Find dy for .
1x3
x2
y
−
=
( )( ) ( )( )
( )
( ) ( )22
2
1x3
2dx
dy
1x3
x62x6
dy
1x3
3x221x3
1x3
x2
ddy
−
−
=∴⇒





−
−−
=
−
−−
=






−
=
dx.byitmultiply
andequationtheofmemberrighttheof
derivativethegetsimplywepractice,In:Note
EXAMPLE 3: Find dy / dx by means of differentials
if xy + sin x = ln y .
( )
( )
( ) ( )
( )
1xy
xcosyy
dx
dy
xcosyy
dx
dy
1xy
xcosyy
dx
dy
dx
dy
xy
dx
dy
xcosyy
dx
dy
xy
dx
1
dydxxcosydxydyxy
dydxxcosydxydyxy
ydy
y
1
dxxcosdxydyx
dy
y
1
dxxcosdxydyx
2
2
2
2
−
+−
=∴
+−=−
−−=−
=++
=++
=++






=++
=++
( ) ( ).tgy,tfxequations
cparametrifor,
dx
yd
and
dx
dy
assuchs,derivative
thefindtoeprocedurtheesinvestigatlessonThis
2
2
==
CHAIN RULE FOR PARAMETRIC EQUATIONS
( ) ( )
dt
dx
dx
dy
dt
d
dx
yd
and
dt
dx
dt
dy
dx
dy
symbols,In
manner.similarain
foundaresderivativeHigher.
dt
dx
to
dt
dy
ofratiotheiscurve
cparametritheon
dx
dy
derivativethethatstatesRuleChainThe
tgyandtfx
equationscparametrithebydefinediscurveaSuppose
RULECHAINHET
2
2 





==
==
Find the derivatives of the following parametric
equations :
tcot
2sint-
2cost
dt
dx
dt
dy
dx
dy
tcos2
dt
dy
andtsin2
dt
dx
:Solution
sint2yt,2cosx.1
−===
=−=
==
3tcot
3sin3t-
3cos3t
dt
dx
dt
dy
dx
dy
t3cos3
dt
dy
andt3sin3
dt
dx
:Solution
3tsiny3t,cosx.2
−===
=−=
==
EXAMPLE :
( ) ( )
( ) ( )
( )
( )
( )2tcot
2tsin
2tcos
dt
dx
dt
dy
dx
dy
2tcos
dt
dy
and2tsin
dt
dx
:Solution
2tsiny,2tcosx.3
+−=
+−
+
==
+=+−=
+=+=
( )
( )
t4sin5tsin2
t4cos5tcos2
t4sin5tsin24
t4cos5tcos24
t4sin20tsin8
t4cos20tcos8
dt
dx
dt
dy
dx
dy
t4cos20tcos8
dt
dy
andt4sin20tsin8
dt
dx
:Solution
5sin4t-t8siny5cos4t,8costx.4
−−
−
=
−−
−
=
−−
−
==
−=
−−=
=+=
2
2
23
dx
yd
find,tty,1txIf.5 +=−=
2
t3
1t2
dt
dx
dt
dy
dx
dy +
==
( )( ) ( )( )
( )
52
2
24
2
2
2
22
2
t9
1t2
dx
yd
t3
1
t9
t61t22t3
dx
yd
dx
dt
t3
1t2
dt
d
dx
yd
+−
=∴
•
+−
=





 +
=
2
2
dx
yd
find,cos41y,sin2xIf.6 θ−=θ=
θ=
θ
θ
=
θ
θ= tan2
cos2
sin4
d
dx
d
dy
dx
dy
( )
θ=∴
θ•θ=
θ
•θ=
θ
θ
θ
=
3
2
2
2
2
2
2
2
2
2
2
sec
dx
yd
secsec
dx
yd
cos2
1
sec2
dx
yd
dx
d
tan2
d
d
dx
yd
( )
( ).0,4attyandt4tx
:curveparametricthetoslinetangenttheFind.7
235
=−=
( ) ( )125tt
2
125tt
2t
t125t
2t
dt
dx
dt
dy
dx
dy
line.tangenttheofslopethegetcanwe
thatso,
dx
dy
findandderivativethefindtohaveWe
22224
−
=
−
=
−
==
( )
( )
( )( )
( )
( )( )
( )
4x
8
1
yx
8
1
4-yislinetangentsecondofequationthethus
8
1
mis0,4atlinetangenttheofslopetheTherefore
8
1
12252
2
dx
dy
,2tat,Now
4x
8
1
yx
8
1
4-yislinetangentofequationthethus
8
1
mis0,4atlinetangenttheofslopetheTherefore
8
1
12252
2
dx
dy
2,tat
definednotis
dx
dy
0,tat
2t0,t
04t,0t
04ttt4t0
becomescurvetheofequationcparametrithe4,0at,Now
2
2
23
2335
+−=→−=→
−=
−=
−−−
=−=
+=→=→
=
=
−
==
=
±==
=−=
=−→−=

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L15 the differentials & parametric equations

  • 2. Consider a function defined by y=f(x) where x is the independent variable. In the four-step rule we introduced the symbol Δx to the denote the increment of x. Now we introduce the symbol dx which we call the differential of x. Similarly, we shall call the symbol dy as the differential of y. To give separate meanings to dx and dy, we shall adopt the following definitions of a function defined by the equation y=f(x). DEFINITION 1: dx = Δx In words, the differential of the independent variable is equal to the increment of the variable.
  • 3. DEFINITION 2: dy = f’ (x) dx In words, the differential of a function is equal to its derivative multiplied by the differential of its independent variable. We emphasize that the differential dx is also an independent variable, it may be assigned any value whatsoever. Therefore, from DEFINITION 2, we see that the differential dy is a function of two independent variables x and dx. It should also be noted that while dx=Δx, dy≠Δy in general. Suppose dx≠0 and we divide both sides of the equation dy = f’ (x) dx
  • 4. by dx. Then we get ( )x'f dx dy = Note that this time dy/dx denotes the quotient of two differentials, dy and dx . Thus the definition of the differential makes it possible to define the derivative of the function as the ratio of two differentials. That is, ( ) xofaldifferentithe yofaldifferentithe dx dy x'f == The differential may be given a geometric interpretation. Consider again the equation y=f(x) and let its graph be as shown below. Let P(x,y) and Q(x+Δx,f(x)+Δx) be two points on the curve. Draw the
  • 5. tangent to the curve at P. Through Q, draw a perpendicular to the x-axis and intersecting the tangent at T. Then draw a line through P, parallel to the x-axis and intersecting the perpendicular through Q at R. Let θ be the inclination of the tangent PT. P Q T R θ
  • 6. From Analytic Geometry, we know that slope of PT = tan θ But triangle PRT, we see that x RT PR RT tan ∆ θ == However, Δx=dx by DEFINITION 1 . Hence dx RT tan =θ But the derivative of y=f(x) at point P is equal to the slope of the tangent line at that same point P. slope of PT = f’(x) Hence, ( ) dx RT x'f =
  • 7. And , RT = f’(x) dx But, dy = f’ (x) dx Hence, RT = dy We see that dy is the increment of the ordinate of the tangent line corresponding to an increment in Δx in x whereas Δy is the corresponding increment of the curve for the same increment in x. We also note that the derivative dy/dx or f’(x) gives the slope of the tangent while the differential dy gives the rise of the tangent line.
  • 8. DIFFERENTIAL FORMULAS Since we have already considered dy/dx as the ratio of two differentials, then the differentiation formulas may now be expressed in terms of differentials by multiplying both sides of the equation by dx. Thus d(c) = 0 d(x) =dx d(cu) = cdu d(u + v) = du + dv d(uv) = udv + vdu d(u/v) = (vdu – udv)/v2 d(un ) = nun-1 du ( ) u2duud =
  • 9. EXAMPLE 1: Find dy for y = x3 + 5 x −1. ( ) ( )dx53xdy dx5dxx3 1x5xddy 2 2 3 += += −+= EXAMPLE 2: Find dy for . 1x3 x2 y − = ( )( ) ( )( ) ( ) ( ) ( )22 2 1x3 2dx dy 1x3 x62x6 dy 1x3 3x221x3 1x3 x2 ddy − − =∴⇒      − −− = − −− =       − = dx.byitmultiply andequationtheofmemberrighttheof derivativethegetsimplywepractice,In:Note
  • 10. EXAMPLE 3: Find dy / dx by means of differentials if xy + sin x = ln y . ( ) ( ) ( ) ( ) ( ) 1xy xcosyy dx dy xcosyy dx dy 1xy xcosyy dx dy dx dy xy dx dy xcosyy dx dy xy dx 1 dydxxcosydxydyxy dydxxcosydxydyxy ydy y 1 dxxcosdxydyx dy y 1 dxxcosdxydyx 2 2 2 2 − +− =∴ +−=− −−=− =++ =++ =++       =++ =++
  • 11. ( ) ( ).tgy,tfxequations cparametrifor, dx yd and dx dy assuchs,derivative thefindtoeprocedurtheesinvestigatlessonThis 2 2 == CHAIN RULE FOR PARAMETRIC EQUATIONS
  • 12. ( ) ( ) dt dx dx dy dt d dx yd and dt dx dt dy dx dy symbols,In manner.similarain foundaresderivativeHigher. dt dx to dt dy ofratiotheiscurve cparametritheon dx dy derivativethethatstatesRuleChainThe tgyandtfx equationscparametrithebydefinediscurveaSuppose RULECHAINHET 2 2       == ==
  • 13. Find the derivatives of the following parametric equations : tcot 2sint- 2cost dt dx dt dy dx dy tcos2 dt dy andtsin2 dt dx :Solution sint2yt,2cosx.1 −=== =−= == 3tcot 3sin3t- 3cos3t dt dx dt dy dx dy t3cos3 dt dy andt3sin3 dt dx :Solution 3tsiny3t,cosx.2 −=== =−= == EXAMPLE :
  • 14. ( ) ( ) ( ) ( ) ( ) ( ) ( )2tcot 2tsin 2tcos dt dx dt dy dx dy 2tcos dt dy and2tsin dt dx :Solution 2tsiny,2tcosx.3 +−= +− + == +=+−= +=+= ( ) ( ) t4sin5tsin2 t4cos5tcos2 t4sin5tsin24 t4cos5tcos24 t4sin20tsin8 t4cos20tcos8 dt dx dt dy dx dy t4cos20tcos8 dt dy andt4sin20tsin8 dt dx :Solution 5sin4t-t8siny5cos4t,8costx.4 −− − = −− − = −− − == −= −−= =+=
  • 15. 2 2 23 dx yd find,tty,1txIf.5 +=−= 2 t3 1t2 dt dx dt dy dx dy + == ( )( ) ( )( ) ( ) 52 2 24 2 2 2 22 2 t9 1t2 dx yd t3 1 t9 t61t22t3 dx yd dx dt t3 1t2 dt d dx yd +− =∴ • +− =       + =
  • 16. 2 2 dx yd find,cos41y,sin2xIf.6 θ−=θ= θ= θ θ = θ θ= tan2 cos2 sin4 d dx d dy dx dy ( ) θ=∴ θ•θ= θ •θ= θ θ θ = 3 2 2 2 2 2 2 2 2 2 2 sec dx yd secsec dx yd cos2 1 sec2 dx yd dx d tan2 d d dx yd
  • 17. ( ) ( ).0,4attyandt4tx :curveparametricthetoslinetangenttheFind.7 235 =−= ( ) ( )125tt 2 125tt 2t t125t 2t dt dx dt dy dx dy line.tangenttheofslopethegetcanwe thatso, dx dy findandderivativethefindtohaveWe 22224 − = − = − ==
  • 18. ( ) ( ) ( )( ) ( ) ( )( ) ( ) 4x 8 1 yx 8 1 4-yislinetangentsecondofequationthethus 8 1 mis0,4atlinetangenttheofslopetheTherefore 8 1 12252 2 dx dy ,2tat,Now 4x 8 1 yx 8 1 4-yislinetangentofequationthethus 8 1 mis0,4atlinetangenttheofslopetheTherefore 8 1 12252 2 dx dy 2,tat definednotis dx dy 0,tat 2t0,t 04t,0t 04ttt4t0 becomescurvetheofequationcparametrithe4,0at,Now 2 2 23 2335 +−=→−=→ −= −= −−− =−= +=→=→ = = − == = ±== =−= =−→−=