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Question
sec(x + y) = xy
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Solution
Given that: sec(x + y) = xy
Differentiating both sides w.r.t. x
`"d"/"dx" sec(x + y) = "d"/"dx"(xy)`
⇒ `sec(x + y) tan(x + y) * "d"/"dx"(x + y) = x*"dy"/"dx" + y*1`
⇒ `sec(x + y)*tan(x + y) (1 + "dy"/"dx") = x*"dy"/"dx" + y`
⇒ `sec(x + y)*tan(x + y) + sec(x + y)*"dy"/"dx"` = y – sec(x + y).tan(x + y)
⇒ `[sec(x + y)* tan(x + y) - x] "dy"/"dx"` = = y – sec(x + y).tan(x + y)
⇒ `"dy"/"dx" = (y - sec(x + y)*tan(x + y))/(sec(x + y)*tan(x + y) - x)`
Hence, `"dy"/"dx" = (y - sec(x + y)*tan(x + y))/(sec(x + y)*tan(x + y) - x)`.
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