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Find dy/dx for the given function: xy = e^(ЁЭСетИТЁЭСж)

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Find `bb(dy/dx)` for the given function:

xy = `e^((x - y))`

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Given, xy = `e^((x - y))`

Taking logarithm of both the sides,

log (xy) = ` log e^((x - y))`

log x + log y = (x − y) loge e  ....[тИ╡ log xy = log x + log y]

log x + log y = x − y ...[тИ╡ loge = 1]

Differentiating both sides with respect to x,

`d/dx log x + d/dx log y = d/dx (x) - d/dx (y)`

`1/x + 1/y dy/dx = 1 - dy/dx `

`1/y dy/dx + dy/dx = 1 - 1/x`

`dy/dx ((1 + y)/y) = 1 - 1/x`

`((1 + y)/y) dy/dx  = (x - 1)/x`

`therefore dy/dx  = (y (x - 1))/(x (1 + y))`

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рдкрд╛рда 5: Continuity and Differentiability - Exercise 5.5 [рдкреГрд╖реНрда резренрео]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдкрд╛рда 5 Continuity and Differentiability
Exercise 5.5 | Q 15 | рдкреГрд╖реНрда резренрео

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