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Logarithmic differentiation
1.
LOGARITHMIC DIFFERENTIATION
2.
Question 1 ππ π
π₯+π¦ = π₯π¦, Show that ππ¦ ππ₯ = π¦(1βπ₯) π₯(π¦β1) π π₯+π¦ = π₯π¦ π₯ + π¦ = log π₯ + log π¦ Differentiate both sides w.r.t x
3.
1 + ππ¦ ππ₯ = 1 π₯ + ππ¦ ππ₯ 1 π¦ ππ¦ ππ₯ 1 β 1 π¦ = 1 π₯ β
1 ππ¦ ππ₯ π¦ β 1 π¦ = 1 β π₯ π₯ ππ¦ ππ₯ = π¦(1 β π₯) π₯(π¦ β 1)
4.
Question 2 ππ π¦
= π₯ π¦ , ππππ£π π‘βππ‘ π₯ ππ¦ ππ₯ = π¦2 1 β π¦ππππ₯ log y = y log x 1 π¦ ππ¦ ππ₯ = π¦ π₯ + ππππ₯ ππ¦ ππ₯
5.
ππ¦ ππ₯ ( 1 π¦ β log π₯
) = π¦ π₯ ππ¦ ππ₯ 1 β π¦ππππ₯ π¦ = π¦ π₯ ππ¦ ππ₯ = π¦2 π₯(1 β π¦ log π₯) π₯ ππ¦ ππ₯ = π¦2 π₯(1 β π¦ππππ₯)
6.
Question 3 Differentiate π₯
π πππ₯ , π₯ > 0 π€. π. π‘ π₯ y = π₯ π πππ₯ log π¦ = π πππ₯ log π₯ 1 π¦ ππ¦ ππ₯ = π πππ₯ π₯ + ππππ₯ πππ π₯
7.
ππ¦ ππ₯ = π¦( π πππ₯ π₯ + ππππ₯
πππ π₯) ππ¦ ππ₯ = π₯ π πππ₯( π πππ₯ π₯ + ππππ₯ πππ π₯) Question 4 π₯ π¦ = π π₯βπ¦ . Prove that ππ¦ ππ₯ = ππππ₯ (ππππ₯π)2
8.
π¦ ππππ₯ =
π₯ β π¦ π¦ π₯ + ππππ₯ ππ¦ ππ₯ = 1 β ππ¦ ππ₯ ππ¦ ππ₯ ππππ₯ + 1 = 1 β π¦ π₯ ππ¦ ππ₯ ππππ₯ + ππππ = π₯ β π¦ π₯
9.
ππ¦ ππ₯ = π₯ β π¦ π₯
log(π₯π) ππ¦ ππ₯ = π¦ππππ₯ π₯ log(π₯π) πππ€ π¦ππππ₯ = π₯ β π¦ π¦ ππππ₯ + π¦ = π₯ -------------1
10.
π¦ ππππ₯ +
1 = π₯ π¦ log π₯π = π₯ π¦ π₯ = 1 log(π₯π) Substituting in 1, ππ¦ ππ₯ = ππππ₯ (ππππ₯π)2 **********************
11.
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