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Question
If y = log `("e"^"x"/"x"^2)`, then `"dy"/"dx" = ?`
Options
`(2 - "x")/"x"`
`("x" - 2)/"x"`
`("e - x")/"ex"`
`("x - e")/"ex"`
Solution
`bb(("x" - 2)/"x")`
Explanation:
y = log `("e"^"x"/"x"^2)`
= log (ex) − log (x2)
= x log e − log x2
y = x − log x2 ...(∵ log e = 1)
Differentiating w.r.t. 'x', we get
`"dy"/"dx"= 1 - 1/("x"^2)."d"/"dx" ("x"^2)`
= `1 - (2"x")/("x"^2)`
= `1 - 2/"x"`
= `("x"- 2)/"x"`
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