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Question
Find `"dy"/"dx"`if, y = `10^("x"^"x") + 10^("x"^10) + 10^(10^"x")`
Solution
y = `10^("x"^"x") + 10^("x"^10) + 10^(10^"x")`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx" (10^("x"^"x") + 10^("x"^10) + 10^(10^"x"))`
`= "d"/"dx" (10^("x"^"x")) + "d"/"dx" (10^("x"^10)) + "d"/"dx" (10^(10^"x"))`
∴ `"dy"/"dx" = 10^("x"^"x") * log 10 * "d"/"dx" ("x"^"x") + 10^("x"^10) * log 10 * "d"/"dx" ("x"^10) + 10^(10^"x") * log 10 * "d"/"dx" (10^"x")`
`= 10^("x"^"x") * log 10 * "x"^"x"(1 + log "x") + 10^("x"^10) * log 10 * 10 "x"^9 + 10^(10^"x") * log 10 * 10^"x" log 10`
∴ `"dy"/"dx" = 10^("x"^"x") * "x"^"x" * log 10(1 + log "x") + 10^("x"^10) * 10 "x"^9 * log 10 + 10^(10^"x") * 10^"x" (log 10)^2`
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