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**Find y _{2} for the following function:**

y = log x + a^{x}

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#### Solution

y = log x + a^{x }

`y_1 = "dy"/"dx" = 1/x + "a"^x log "a"` ....`[because "d"/"dx" ("a"^x) = "a"^x log "a"]`

`y_2 = ("d"^2y)/"dx"^2 = "d"/"dx"(1/x) + log "a" "d"/"dx" ("a"^x)`

`= (-1)/x^2 + (log "a")("a"^x log "a")`

`= (-1)/x^2 + ("a"^x log "a")^2`

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