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प्रश्न
Find `"dy"/"dx"` if, y = log(ax2 + bx + c)
उत्तर
y = log(ax2 + bx + c)
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx"` [log(ax2 + bx + c)]
`= 1/("ax"^2 + "bx" + "c") * "d"/"dx" ("ax"^2 + "bx" + "c")`
`= 1/("ax"^2 + "bx" + "c") * ["a"("2x") + "b" + 0]`
∴ `"dy"/"dx" = ("2ax" + "b")/("ax"^2 + "bx" + "c")`
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