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प्रश्न
Find `"dy"/"dx"`if, y = `"x"^("x"^"2x")`
उत्तर
y = `"x"^("x"^"2x")`
Taking logarithm of both sides, we get
log y = log`("x")^("x"^"2x")`
∴ log y = `"x"^"2x" * log "x"`
Differentiating both sides w.r.t.x, we get
`1/"y" * "dy"/"dx" = "x"^"2x" * "d"/"dx" (log "x") + log "x" * "d"/"dx"("x"^"2x")`
∴ `1/"y"*"dy"/"dx" = "x"^"2x" * 1/"x" + log "x" * "d"/"dx"("x"^"2x")` .....(i)
Let u = `"x"^"2x"`
Taking logarithm of both sides, we get
log u = `log "x"^"2x" = "2x" * log"x"`
Differentiating both sides w.r.t.x, we get
`1/"u" * "du"/"dx" = "2x" * "d"/"dx" (log "x") + log "x" * "d"/"dx"(2"x")`
∴ `1/"u" * "du"/"dx" = "2x" * 1/"x" + log "x" * (2)`
∴ `1/"u" * "du"/"dx"` = 2 + 2 log x
∴ `"du"/"dx"` = u(2 + 2 log x)
∴ `"du"/"dx"` = 2u(1 + log x)
∴ `"du"/"dx" = "2x"^"2x" (1 + log "x")` ....(ii)
Substituting (ii) in (i), we get
`1/"y" * "dy"/"dx" = "x"^(2"x") * 1/"x" + (log "x")(2"x"^"2x")`(1 + log x)
∴ `"dy"/"dx" = "y"["x"^"2x"/"x" + 2"x"^(2"x") * log "x"(1 + log "x")]`
∴ `"dy"/"dx" = "x"^("x"^"2x") * "x"^"2x" log "x"[1/("x log x") + 2 (1 + log "x")]`
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