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प्रश्न
Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`
उत्तर
`"y" = "x"^("e"^"x")`
Taking logarithm of both the sides,
log y = ex log x
Differentiating both sides w.r.t. x,
`1/"y" "dy"/"dx" = "e"^"x" . "d"/"dx"(log"x") + "log x" ."d"/"dx" ("e"^"x")`
`= "e"^"x". 1/"x" + "log x". "e"^"x"`
`= "e"^"x"(1/"x" + "log x")`
`therefore "dy"/"dx" = "y".["e"^"x" (1/"x" + "log x")]`
`= "x"^("e"^"x") . "e"^"x" (1/"x" + "log x")`
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