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Question
Find `("d"^2"y")/"dx"^2`, if y = `"x"^2 * "e"^"x"`
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Solution
y = `"x"^2 * "e"^"x"`
Differentiating both sides w.r.t. t, we get
`"dy"/"dx" = "x"^2 * "d"/"dx" ("e"^"x") * "d"/"dx" ("x"^2)`
`= "x"^2 * "e"^"x" + "e"^"x" ("2x")`
`"dy"/"dx" = ("x"^2 + 2"x") * "e"^"x"`
Again, differentiating both sides w.r.t. x, we get
`("d"^2"y")/"dx"^2 = ("x"^2 + 2"x") * "d"/"dx" ("e"^"x") + "e"^"x" * "d"/"dx" ("x"^2 + 2"x")`
`= ("x"^2 + 2"x") * "e"^"x" + "e"^"x"`(2x + 2)
`= "e"^"x"("x"^2 + 2"x" + 2"x" + 2)`
∴ `("d"^2"y")/"dx"^2 = "e"^"x"("x"^2 + 4"x" + 2)`
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