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प्रश्न
sinn (ax2 + bx + c)
उत्तर
Let y = sinn (ax2 + bx + c)
Differentiating both sides w.r.t. x
`"dy"/"dx" = "d"/"dx" sin^"n" ("a"x^2 + "b"x + "c")`
= `"n" * sin^("n" - 1) ("a"x^2 + "b"x + "c") * "d"/"dx" sin("a"x^2 + "b"x + "c")`
= `"n" * sin^("n" - 1) ("a"x^2 + "b"x + "c") * cos("a"x^2 + "b"x + "c") * "d"/"dx" ("a"x^2 + "b"x + "c")`
= `"n" * sin^("n" - 1) ("a"x^2 + "b"x + "c") * cos("a"x^2 + "b"x + "c") * (2"a"x + "b")`
Hence, `"dy"/"dx" = "n" (2"a"x + "b") * sin^("n" - 1)("a"x^2 + "b"x + "c")*cos("a"x^2 + "b"x + "c")`
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