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Question
Find `("d"^2"y")/"dx"^2`, if y = `"e"^((2"x" + 1))`.
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Solution
y = `"e"^((2"x" + 1))`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "e"^((2"x" + 1)) * "d"/"dx" (2"x" + 1)`
`"dy"/"dx" = "e"^((2"x" + 1)) * (2 + 0)`
`"dy"/"dx" = 2"e"^((2"x" + 1))`
Again, differentiating both sides w.r.t. x, we get
`("d"^2"y")/"dx"^2 = 2 * "d"/"dx" "e"^((2"x" + 1))`
`= 2"e"^((2"x" + 1)) * "d"/"dx" (2"x" + 1)`
`= 2"e"^((2"x" + 1)) * (2 + 0)`
∴ `("d"^2"y")/"dx"^2 = 4"e"^((2"x" + 1))`
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