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Question
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Solution
Given 4x + 3y = log(4x – 3y)
Differentiating both sides with respect to x,
`4(1) + 3"dy"/"dx" = 1/((4x - 3y)) "d"/"dx" (4x - 3y)`
`4 + 3"dy"/"dx" = 1/((4x - 3y)) (4(1) - 3"dy"/"dx")`
`(4x - 3y)(4 + 3 "dy"/"dx") = 4 - 3 "dy"/"dx"`
`16x + 12x "dy"/"dx" - 12y - 9y "dy"/"dx" = 4 - 3 "dy"/"dx"`
`12x "dy"/"dx" + 3 "dy"/"dx" - 9y "dy"/"dx" = 4 - 16x + 12y`
`"dy"/"dx"`[12x + 3 - 9y] = 4[1 - 4x + 3y]
`"dy"/"dx" = (4[1 - 4x + 3y])/(3[4x + 1 - 3y])`
= `(4[1 - 4x + 3y])/(3[1 + 4x - 3y])`
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