Logarithmic differentiation
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This document contains solutions to 4 logarithmic differentiation problems: 1) It shows that dy/dx = y(1-x)/(x(y-1)) if e^(x+y) = xy. 2) It proves that x(dy/dx) = y^2/(1 - ylogx) if y = x^y. 3) It finds the derivative of x sinx with respect to x. 4) It proves that dy/dx = logx/(logxe)^2 if x^y = e^(x-y).
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Logarithmic differentiation
- 2. Question 1 𝑖𝑓 𝑒 𝑥+𝑦 = 𝑥𝑦, Show that 𝑑𝑦 𝑑𝑥 = 𝑦(1−𝑥) 𝑥(𝑦−1) 𝑒 𝑥+𝑦 = 𝑥𝑦 𝑥 + 𝑦 = log 𝑥 + log 𝑦 Differentiate both sides w.r.t x
- 3. 1 + 𝑑𝑦 𝑑𝑥 = 1 𝑥 + 𝑑𝑦 𝑑𝑥 1 𝑦 𝑑𝑦 𝑑𝑥 1 − 1 𝑦 = 1 𝑥 − 1 𝑑𝑦 𝑑𝑥 𝑦 − 1 𝑦 = 1 − 𝑥 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦(1 − 𝑥) 𝑥(𝑦 − 1)
- 4. Question 2 𝑖𝑓 𝑦 = 𝑥 𝑦 , 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦2 1 − 𝑦𝑙𝑜𝑔𝑥 log y = y log x 1 𝑦 𝑑𝑦 𝑑𝑥 = 𝑦 𝑥 + 𝑙𝑜𝑔𝑥 𝑑𝑦 𝑑𝑥
- 5. 𝑑𝑦 𝑑𝑥 ( 1 𝑦 − log 𝑥 ) = 𝑦 𝑥 𝑑𝑦 𝑑𝑥 1 − 𝑦𝑙𝑜𝑔𝑥 𝑦 = 𝑦 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦2 𝑥(1 − 𝑦 log 𝑥) 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦2 𝑥(1 − 𝑦𝑙𝑜𝑔𝑥)
- 6. Question 3 Differentiate 𝑥 𝑠𝑖𝑛𝑥 , 𝑥 > 0 𝑤. 𝑟. 𝑡 𝑥 y = 𝑥 𝑠𝑖𝑛𝑥 log 𝑦 = 𝑠𝑖𝑛𝑥 log 𝑥 1 𝑦 𝑑𝑦 𝑑𝑥 = 𝑠𝑖𝑛𝑥 𝑥 + 𝑙𝑜𝑔𝑥 𝑐𝑜𝑠𝑥
- 7. 𝑑𝑦 𝑑𝑥 = 𝑦( 𝑠𝑖𝑛𝑥 𝑥 + 𝑙𝑜𝑔𝑥 𝑐𝑜𝑠𝑥) 𝑑𝑦 𝑑𝑥 = 𝑥 𝑠𝑖𝑛𝑥( 𝑠𝑖𝑛𝑥 𝑥 + 𝑙𝑜𝑔𝑥 𝑐𝑜𝑠𝑥) Question 4 𝑥 𝑦 = 𝑒 𝑥−𝑦 . Prove that 𝑑𝑦 𝑑𝑥 = 𝑙𝑜𝑔𝑥 (𝑙𝑜𝑔𝑥𝑒)2
- 8. 𝑦 𝑙𝑜𝑔𝑥 = 𝑥 − 𝑦 𝑦 𝑥 + 𝑙𝑜𝑔𝑥 𝑑𝑦 𝑑𝑥 = 1 − 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑙𝑜𝑔𝑥 + 1 = 1 − 𝑦 𝑥 𝑑𝑦 𝑑𝑥 𝑙𝑜𝑔𝑥 + 𝑙𝑜𝑔𝑒 = 𝑥 − 𝑦 𝑥
- 9. 𝑑𝑦 𝑑𝑥 = 𝑥 − 𝑦 𝑥 log(𝑥𝑒) 𝑑𝑦 𝑑𝑥 = 𝑦𝑙𝑜𝑔𝑥 𝑥 log(𝑥𝑒) 𝑛𝑜𝑤 𝑦𝑙𝑜𝑔𝑥 = 𝑥 − 𝑦 𝑦 𝑙𝑜𝑔𝑥 + 𝑦 = 𝑥 -------------1
- 10. 𝑦 𝑙𝑜𝑔𝑥 + 1 = 𝑥 𝑦 log 𝑥𝑒 = 𝑥 𝑦 𝑥 = 1 log(𝑥𝑒) Substituting in 1, 𝑑𝑦 𝑑𝑥 = 𝑙𝑜𝑔𝑥 (𝑙𝑜𝑔𝑥𝑒)2 **********************
- 11. For more videos on grade 12 differentiation subscribe to the following channel UCjmCXXIjd03JQad8-rUzs0Q Youtube channel id