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Question
Find y2 for the following function:
y = log x + ax
Solution
y = log x + ax
`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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