Sec 3.7 Differentials
Δx = Increment in x := Change in the value of x
Δy = Increment in y = f (x) := Change in the value of f (x) when there
is a change in x
Q. Let f (x) = x2 . Find Δx and Δy when x changes from
(a) 1 to 1.1
(b) 2 to 1.98
Soln. (a) Δx = 1.1 − 1 = 0.1
Δf (x) = f (1.1) − f (1) = (1.1)2 − (1)2 = 1.21 − 1 = 0.21
(b) Δx = 1.98 − 2 = − 0.02
Δf (x) = (1.98)2 − (2)2 = 3.9204 − 4 = − 0.0796
The differential dx of the independent variable x is defined to
be Δx
The differential dy of the dependent variable y = f (x) , is defined
to be dy := f ′(x) (dx) .
The value of dy is numerically equal to f ′(x) (Δx) , since dx = Δx .
Q Let y = x2 . Find dx and dy when x changes from 2 to 1.98 .
Soln. dy = (2x)(dx) .
dx = 1.98 − 2 = − 0.02, x = 2 , so dy = 2(2)(− 0.02) = − 0.08 .
Q Let y = x2 . Find dx and dy when x changes from 2 to 1.98 .
Soln. dy = (2x)(dx) .
dx = 1.98 − 2 = − 0.02, x = 2 ,
so dy = 2(2)(− 0.02) = − 0.08 .
Δy = 1.982 − 4 = − 0.0796 , and we found dy = − 0.08
dy ≃ Δy dy = f ′(x)dx is an approximation to the
actual change in y when
x changes by
dx ​
• dy is easier to calculate
• dx is equal to the actual change in x that is Δx .
Let us look at some examples more examples of differentials.
y=
x−1
x2+1 2
x−1)(2x)
, dy = (x +1) − (
dx =
(x2+1)2
−x2+2x+1
dx
(x2+1)2
y = (2x2 + 3)1/3 ,
dy = (1/3)(2x2 + 3)−2/3 (4x) dx y = x + 2x dy = (1 − x22 )dx
,
Q. Approximate the value of √26.5 using differentials.
Soln. Δf (x) = f (x + Δx) − f (x) f (x + Δx) = f (x) + Δf (x) f (x + Δx) ≃ f (x) + df (x) f (x + Δx) ≃ f (x) + f ′(x) dx If we set f (x) = √x , and x = 25 , and Δx = 1.5 , we want f (25 + 1.5) .
f (25 + 1.5) ≃ f (25) + f ′(25)(1.5) 1
√25 + 1.5 ≃ √25 + 2√25
(1.5)
√26.5 ≃ 5 + (1.5)/10 = 5.15 • It is much easier to do the above calculation compared to
finding √26.5 .
Let us look at some more uses of this technique.
√10 = √9 + 1 ≃ √9 + 2√19 (1) , f (x) = √x , x = 9, dx = 1
√3 7.8 = √3 8 − 0.2 ≃ √3 8 + 1 2/3 (− 0.2) , f (x) = √3 x , x = 8 , dx = − 0.2
3(8)
1
√0.089 = √0.09 − 0.001 ≃ √0.09 + 2(0.3)
(− 0.001)
,
f (x) = √x , x = 0.09 , dx = − 0.001
1
√3 0.00096 = √3 0.001 − 0.00004 ≃ √3 0.001 + 3(0.001)
2/3 (− 0.00004) ,
f (x) = √3 x , x = 0.001 , dx = − 0.00004
1
Q. Approximate √4.02 + √4.02
1
1
Soln. √4.02 + √4.02
= √4 + 0.02 + √4+0.02
f (x) = x1/2 + x−1/2 ,x
= 4, dx=0.02
df (x) = ( 2x11/2 − 2x13/2 )dx
1
1
f (4.02) = f (4) + f ′(4)(0.02) = 2 + 2−1 + ( 2(2)
− 2(8)
)(0.02)
f (4.02) = 2 + 0.5 + (0.25 − 0.0625)(0.02) = 2.5 + (0.1875)(0.02) = 2.5 + 0.003750 f (4.02) = 2.503750 .
Please note Δy = f (x + Δx) − f (x) is the actual change in f (x) when x
changes by Δx , while dy = f ′(x)dx is an approximation to Δy
Q. 36 The supply equation for a certain brand of radio is given
by p = s(x) = 0.3√x + 10 . where x is the quantity supplied in
thousands and p is the unit price in dollars. Use differentials
to approximate the change in price when the quantity supplied is
increased from 10000 units to 10500 units.
Soln.
x=10, Δx = dx = 0.5
Δp ≃ dp = s′(x)dx = 20.3
dx =
√x
0.3
(0.5)
2√10
= 0.0237
The approximate change in price is $ 0.0237 .
Estimating errors in measurement.
Q. Suppose the side of a cube is measured with a maximum
percentage error of 2%. Use differentials to estimate the maximum
percentage error in the ​calculated​ volume of the cube.
Soln. V (x) = x3 . Let Δx denote the error in measurement. It is
given that the maximum value of Δx
x = 0.02 . we want to find the
maximum value of
ΔV (x)
V (x)
≃
3x2dx
x3
=
3dx
x
= 3(0.02) = 0.06 . The maximum
percentage error in the calculated volume is ​approximately​ 6%.
Q. A coat of paint of thickness 0.05 cm is to be applied
uniformly to the faces of a cube of edge 30cm. Use differentials
to find the approximate amount of paint required. For the job.
Soln. After the paint is applied the edge of the cube will change
from 30cm to 30.1cm. Let x denote the edge of the cube, and V (x)
be the corresponding volume. Then V (x) = x3, x = 30, dx = 0.1 and
dV = 3x2 dx = 3(30)2(0.1) = 3(900)(0.1) = 270cm3 . So the volume of the
cube changes by approximately 270cm3 .