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LOGARITHMIC
DIFFERENTIATION
Question 1
𝑖𝑓 𝑒 π‘₯+𝑦
= π‘₯𝑦,
Show that
𝑑𝑦
𝑑π‘₯
=
𝑦(1βˆ’π‘₯)
π‘₯(π‘¦βˆ’1)
𝑒 π‘₯+𝑦
= π‘₯𝑦
π‘₯ + 𝑦 = log π‘₯ + log 𝑦
Differentiate both sides w.r.t x
1 +
𝑑𝑦
𝑑π‘₯
=
1
π‘₯
+
𝑑𝑦
𝑑π‘₯
1
𝑦
𝑑𝑦
𝑑π‘₯
1 βˆ’
1
𝑦
=
1
π‘₯
βˆ’ 1
𝑑𝑦
𝑑π‘₯
𝑦 βˆ’ 1
𝑦
=
1 βˆ’ π‘₯
π‘₯
𝑑𝑦
𝑑π‘₯
=
𝑦(1 βˆ’ π‘₯)
π‘₯(𝑦 βˆ’ 1)
Question 2
𝑖𝑓 𝑦 = π‘₯ 𝑦
, π‘π‘Ÿπ‘œπ‘£π‘’ π‘‘β„Žπ‘Žπ‘‘ π‘₯
𝑑𝑦
𝑑π‘₯
=
𝑦2
1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯
log y = y log x
1
𝑦
𝑑𝑦
𝑑π‘₯
=
𝑦
π‘₯
+ π‘™π‘œπ‘”π‘₯
𝑑𝑦
𝑑π‘₯
𝑑𝑦
𝑑π‘₯
(
1
𝑦
βˆ’ log π‘₯ ) =
𝑦
π‘₯
𝑑𝑦
𝑑π‘₯
1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯
𝑦
=
𝑦
π‘₯
𝑑𝑦
𝑑π‘₯
=
𝑦2
π‘₯(1 βˆ’ 𝑦 log π‘₯)
π‘₯
𝑑𝑦
𝑑π‘₯
=
𝑦2
π‘₯(1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯)
Question 3
Differentiate π‘₯ 𝑠𝑖𝑛π‘₯
, π‘₯ > 0 𝑀. π‘Ÿ. 𝑑 π‘₯
y = π‘₯ 𝑠𝑖𝑛π‘₯
log 𝑦 = 𝑠𝑖𝑛π‘₯ log π‘₯
1
𝑦
𝑑𝑦
𝑑π‘₯
=
𝑠𝑖𝑛π‘₯
π‘₯
+ π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯
𝑑𝑦
𝑑π‘₯
= 𝑦(
𝑠𝑖𝑛π‘₯
π‘₯
+ π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯)
𝑑𝑦
𝑑π‘₯
= π‘₯ 𝑠𝑖𝑛π‘₯(
𝑠𝑖𝑛π‘₯
π‘₯
+ π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯)
Question 4
π‘₯ 𝑦
= 𝑒 π‘₯βˆ’π‘¦
. Prove that
𝑑𝑦
𝑑π‘₯
=
π‘™π‘œπ‘”π‘₯
(π‘™π‘œπ‘”π‘₯𝑒)2
𝑦 π‘™π‘œπ‘”π‘₯ = π‘₯ βˆ’ 𝑦
𝑦
π‘₯
+ π‘™π‘œπ‘”π‘₯
𝑑𝑦
𝑑π‘₯
= 1 βˆ’
𝑑𝑦
𝑑π‘₯
𝑑𝑦
𝑑π‘₯
π‘™π‘œπ‘”π‘₯ + 1 = 1 βˆ’
𝑦
π‘₯
𝑑𝑦
𝑑π‘₯
π‘™π‘œπ‘”π‘₯ + π‘™π‘œπ‘”π‘’ =
π‘₯ βˆ’ 𝑦
π‘₯
𝑑𝑦
𝑑π‘₯
=
π‘₯ βˆ’ 𝑦
π‘₯ log(π‘₯𝑒)
𝑑𝑦
𝑑π‘₯
=
π‘¦π‘™π‘œπ‘”π‘₯
π‘₯ log(π‘₯𝑒)
π‘›π‘œπ‘€ π‘¦π‘™π‘œπ‘”π‘₯ = π‘₯ βˆ’ 𝑦
𝑦 π‘™π‘œπ‘”π‘₯ + 𝑦 = π‘₯
-------------1
𝑦 π‘™π‘œπ‘”π‘₯ + 1 = π‘₯
𝑦 log π‘₯𝑒 = π‘₯
𝑦
π‘₯
=
1
log(π‘₯𝑒)
Substituting in 1,
𝑑𝑦
𝑑π‘₯
=
π‘™π‘œπ‘”π‘₯
(π‘™π‘œπ‘”π‘₯𝑒)2
**********************
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Logarithmic differentiation

  • 2. Question 1 𝑖𝑓 𝑒 π‘₯+𝑦 = π‘₯𝑦, Show that 𝑑𝑦 𝑑π‘₯ = 𝑦(1βˆ’π‘₯) π‘₯(π‘¦βˆ’1) 𝑒 π‘₯+𝑦 = π‘₯𝑦 π‘₯ + 𝑦 = log π‘₯ + log 𝑦 Differentiate both sides w.r.t x
  • 3. 1 + 𝑑𝑦 𝑑π‘₯ = 1 π‘₯ + 𝑑𝑦 𝑑π‘₯ 1 𝑦 𝑑𝑦 𝑑π‘₯ 1 βˆ’ 1 𝑦 = 1 π‘₯ βˆ’ 1 𝑑𝑦 𝑑π‘₯ 𝑦 βˆ’ 1 𝑦 = 1 βˆ’ π‘₯ π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦(1 βˆ’ π‘₯) π‘₯(𝑦 βˆ’ 1)
  • 4. Question 2 𝑖𝑓 𝑦 = π‘₯ 𝑦 , π‘π‘Ÿπ‘œπ‘£π‘’ π‘‘β„Žπ‘Žπ‘‘ π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦2 1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯ log y = y log x 1 𝑦 𝑑𝑦 𝑑π‘₯ = 𝑦 π‘₯ + π‘™π‘œπ‘”π‘₯ 𝑑𝑦 𝑑π‘₯
  • 5. 𝑑𝑦 𝑑π‘₯ ( 1 𝑦 βˆ’ log π‘₯ ) = 𝑦 π‘₯ 𝑑𝑦 𝑑π‘₯ 1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯ 𝑦 = 𝑦 π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦2 π‘₯(1 βˆ’ 𝑦 log π‘₯) π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦2 π‘₯(1 βˆ’ π‘¦π‘™π‘œπ‘”π‘₯)
  • 6. Question 3 Differentiate π‘₯ 𝑠𝑖𝑛π‘₯ , π‘₯ > 0 𝑀. π‘Ÿ. 𝑑 π‘₯ y = π‘₯ 𝑠𝑖𝑛π‘₯ log 𝑦 = 𝑠𝑖𝑛π‘₯ log π‘₯ 1 𝑦 𝑑𝑦 𝑑π‘₯ = 𝑠𝑖𝑛π‘₯ π‘₯ + π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯
  • 7. 𝑑𝑦 𝑑π‘₯ = 𝑦( 𝑠𝑖𝑛π‘₯ π‘₯ + π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯) 𝑑𝑦 𝑑π‘₯ = π‘₯ 𝑠𝑖𝑛π‘₯( 𝑠𝑖𝑛π‘₯ π‘₯ + π‘™π‘œπ‘”π‘₯ π‘π‘œπ‘ π‘₯) Question 4 π‘₯ 𝑦 = 𝑒 π‘₯βˆ’π‘¦ . Prove that 𝑑𝑦 𝑑π‘₯ = π‘™π‘œπ‘”π‘₯ (π‘™π‘œπ‘”π‘₯𝑒)2
  • 8. 𝑦 π‘™π‘œπ‘”π‘₯ = π‘₯ βˆ’ 𝑦 𝑦 π‘₯ + π‘™π‘œπ‘”π‘₯ 𝑑𝑦 𝑑π‘₯ = 1 βˆ’ 𝑑𝑦 𝑑π‘₯ 𝑑𝑦 𝑑π‘₯ π‘™π‘œπ‘”π‘₯ + 1 = 1 βˆ’ 𝑦 π‘₯ 𝑑𝑦 𝑑π‘₯ π‘™π‘œπ‘”π‘₯ + π‘™π‘œπ‘”π‘’ = π‘₯ βˆ’ 𝑦 π‘₯
  • 9. 𝑑𝑦 𝑑π‘₯ = π‘₯ βˆ’ 𝑦 π‘₯ log(π‘₯𝑒) 𝑑𝑦 𝑑π‘₯ = π‘¦π‘™π‘œπ‘”π‘₯ π‘₯ log(π‘₯𝑒) π‘›π‘œπ‘€ π‘¦π‘™π‘œπ‘”π‘₯ = π‘₯ βˆ’ 𝑦 𝑦 π‘™π‘œπ‘”π‘₯ + 𝑦 = π‘₯ -------------1
  • 10. 𝑦 π‘™π‘œπ‘”π‘₯ + 1 = π‘₯ 𝑦 log π‘₯𝑒 = π‘₯ 𝑦 π‘₯ = 1 log(π‘₯𝑒) Substituting in 1, 𝑑𝑦 𝑑π‘₯ = π‘™π‘œπ‘”π‘₯ (π‘™π‘œπ‘”π‘₯𝑒)2 **********************
  • 11. For more videos on grade 12 differentiation subscribe to the following channel UCjmCXXIjd03JQad8-rUzs0Q Youtube channel id