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प्रश्न
If sin y = x sin (a + y), then show that `"dy"/"dx" = (sin^2(a + y))/(sina)`.
योग
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उत्तर
sin y = x sin(a + y)
⇒ `x = sin y/(sin (a + y))` ....(i)
Differentiating (i) w.r.t.x,
⇒ 1 = `(sin(a + y).(d/dx sin y) - sin y. (d/dx sin (a + y)))/sin^2 (a + y)`
⇒ ` sin(a + y).cos y - d/dx - sin y. cos (a + y). d/dx = sin^2 (a + y)`]
⇒ `d/dx [ sin ( a + y) . cos y - sin y. cos ( a + y)] = sin^2 (a + y)`
⇒ `dy/dx[ sin ( a + y - y)] = sin^2 (a + y)`
⇒ `dy/dx = (sin^2 (a + y))/(sin a)`
Hence proved.
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