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If ЁЭСе =ЁЭСТ^ЁЭСе/ЁЭСж, then show that dy/dx =ЁЭСетИТЁЭСж/ЁЭСетБвlogтБбЁЭСе

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If x = `e^(x/y)`, then show that `dy/dx = (x - y)/(xlogx)`

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x = `e^(x/y)`

∴ `x/y` = log x    ...(1)

∴ y = `x/logx`

∴ `dy/dx = d/dx(x/log x)`

= `((log x) * d/dx(x) - x * d/dx(log x))/((log x)`

= `((log x) xx 1 - x xx (1)/x)/((log x)^2`

= `(log x - 1)/((log x)(log x)`

= `(x/y - 1)/((x/y)(log x)`    ...[By (1)]

= `(x - y)/(x log x)`

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рдЕрдзреНрдпрд╛рдп 1: Differentiation - Miscellaneous Exercise 1 (II) [рдкреГрд╖реНрда ремрек]

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рдмрд╛рд▓рднрд╛рд░рддреА Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
рдЕрдзреНрдпрд╛рдп 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 5.5 | рдкреГрд╖реНрда ремрек
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