1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard VII
Mathematics
Power of a Power
Using differe...
Question
Using differentiation find approximate value of
(
26
)
1
3
Open in App
Solution
Let
y
=
(
x
)
2
3
Here
x
=
26
Δ
x
=
−
1
Differentiating w.r.t x
d
y
d
x
=
1
3
x
1
/
3
−
1
=
1
3
x
−
2
/
3
=
1
3
x
2
/
3
Δ
y
=
d
y
d
x
Δ
x
=
1
3
x
2
/
3
Δ
x
Putting values,
Δ
y
=
1
3
x
2
/
3
×
−
1
=
−
1
3
×
3
2
=
−
1
3
×
9
=
−
1
27
Δ
y
=
−
0.037037
We know,
Δ
y
=
6
(
x
+
Δ
x
)
−
6
(
x
)
=
(
x
+
Δ
x
)
1
/
3
−
x
1
/
3
Putting values,
−
0.037037
=
(
27
+
(
−
1
)
)
1
/
3
−
(
27
)
1
/
3
=
26
1
/
3
−
(
27
)
1
/
3
=
26
1
/
3
−
3
−
0.037037
+
3
=
26
1
/
3
2.9629
=
26
1
/
2
∴
Approximate value of
26
1
/
3
is
2.9629
.
Suggest Corrections
0
Similar questions
Q.
Using differentials, the approximate value of
(
26
)
1
/
3
is
2.9629
Q.
Using differentials, find the approximate value of the following up to 3 places of decimal.
(
26
)
1
3
Q.
Differentiate and find the approximate value of
(
26
)
1
3
Q.
Use differentiation to approximate
(
26
)
1
3
.
Q.
Using differentials, find the sum of digits approximate value of the following up to
3
places of decimal.
(
26
)
1
3
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
Laws of Exponents
MATHEMATICS
Watch in App
Explore more
Power of a Power
Standard VII Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Solve
Textbooks
Question Papers
Install app