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प्रश्न
Find the derivatives of the following:
If sin y = x sin(a + y), the prove that `("d"y)/("d"x) = (sin^2("a" + y))/sin"a"`, a ≠ nπ
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उत्तर
Given sin y = x sin(a + y) ........(1)
Differentiating with respect to x, we get
`cos ("d"y)/("d"x) = x cos("a" + y) (0 + ("d"y)/("d"x)) + sin("a" + y) * 1`
`cos y ("d"y)/("d"x) = xcos("a" + y) ("d"y)/("d"x) + sin("a" + y)`
`cos y ("d"y)/("d"x) - xcos("a" + y) ("d"y)/("d"x) = sin("a" + y)`
`("d"y)/("d"x) = (sin("a" + y))/(cosy- xcos("a" + y))` ........(2)
From equation (1) we have, x = `sin y/(sin("a" + y))`
Substituting for x in equation (2) we get
`("d"y)/("d"x) = (sin("a" + y))/(cosy - siny/(sin("a" + y)) * cos("a" + y))`
`("d"y)/("d"x) = (sin("a" + y))/((sin("a" + y) cosy - cos("a" + y) sin y)/(sin("a" + y))`
= `(sin^2("a" + y))/(sin ["a" + y - y])`
sin(A – B) = sinA cosB – cosA sin B
`("d"y)/("d"x) = (sin^2("a" + y))/sin "a"`
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