
4-26 Calculus – Differentiation and Integration
So that Dx = dx = 2, y = 5 and 127
3
= y + Dy
We have
dy x dx
x x
=
=
=
=
−
−
−
1
3
1
3
1
3
125 2
2
75
2 3
2 3
2 3
/
/
/
( ) ( )
.
Therefore,
127
5
2
75
377
75
3
= +
≈ +
= +
=
=
y y
y dy
b. We assume that y = tan x, where x = p/4.
So that Dx = dx = 0.89 - (p/4) = 0.1046, y = 1 and tan (0.89) = y + Dy.
We have
dy x dx
x x
=
=
=
( )
=
sec
sec
sec .
( .
2
2
2
4
0 1046
1 999
p
99 0 1046
0 20919
)( . )
. .=
Therefore,
tan( . )
.
. .
0 89
1 0 20919
1 20919
= +
≈ +
= +
=
y y
y dy
c. We assume that y = ln x, where x = 1.