Question

using differentials find the value of (0.73) its power is 1/3

Answer 1

Toolbox:
Let
y=f(x)
y=f(x)
Δx
Δx
denote a small increment in
x
x
Δy=f(x+Δx)−f(x)
Δy=f(x+Δx)−f(x)
dy=(
dy
dx
)Δx
dy=(dydx)Δx
Step 1:
Let
y=
x
1
3
y=x13
Let
x=0.008
x=0.008
dx=0.001
dx=0.001
So that
x+dx=0.009
x+dx=0.009
Now
(x+Δx
)
1
3
−
x
1
3
(x+Δx)13−x13
⇒(0.009
)
1
3
−(0.008
)
1
3
⇒(0.009)13−(0.008)13
⇒(0.009
)
1
3
−0.2
⇒(0.009)13−0.2
∴(0.009
)
1
3
−0.2+Δy
∴(0.009)13−0.2+Δy
-----(1)
Step 2:
Also
dy
dx
dydx
Δx
Δx
is approximately equal to
dy
dy
dy=(
dy
dx
)
dy=(dydx)
Δx
Δx
=
1
3
x
2
3
=13x23
Δx
Δx
=
1
3(0.008
)
2
3
=13(0.008)23
×0.001
×0.001
=
1
3(0.2
)
2
=13(0.2)2
×0.001
×0.001
=
0.001
3×0.04
=0.0013×0.04
=
0.001
0.12
=0.0010.12
=
1
120
=1120
=0.008
=0.008
Step 3:
Approximate value of
Δy=dy=0.008
Δy=dy=0.008
Hence from Equation(1) we have
(0.009
)
1
3
=0.2+0.008
(0.009)13=0.2+0.008
=0.208