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Question
Find `("d"^2"y")/"dx"^2`, if y = `sqrt"x"`
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Solution
y = `sqrt"x"`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = 1/(2sqrt"x")`
∴ `"dy"/"dx" = 1/2 "x"^(-1/2)`
Again, differentiating both sides w.r.t. x , we get
`("d"^2"y")/"dx"^2 = 1/2 * "d"/"dx"("x"^(-1/2))`
`= 1/2 (- 1/2)* "x"^(- 3/2)`
∴ `("d"^2"y")/"dx"^2 = (-1)/4 "x"^(-3/2)`
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