हिंदी

Find dydxd2ydx2, if y = exe(2x+1).

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प्रश्न

Find `("d"^2"y")/"dx"^2`, if y = `"e"^((2"x" + 1))`.

योग
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उत्तर

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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अध्याय 3: Differentiation - EXERCISE 3.6 [पृष्ठ ९८]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 3 Differentiation
EXERCISE 3.6 | Q 2. 2) | पृष्ठ ९८

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