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If y = sin-1[acosx-bsinxa2+b2], then find dydx

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प्रश्न

If y = `sin^-1[("a"cosx - "b"sinx)/sqrt("a"^2 + "b"^2)]`, then find `("d"y)/("d"x)`

योग
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उत्तर

y = `sin^-1[("a"cosx - "b"sinx)/sqrt("a"^2 + "b"^2)]`

= `sin^-1["a"/sqrt("a"^2 + "b"^2) cosx - "b"/sqrt("a"^2 + "b"^2) sinx]`

Put `"a"/sqrt("a"^2 + "b"^2)` = sin t and `"b"/sqrt("a"^2 + "b"^2)` = cos t

Also, sin2t + cos2t = `("a"^2)/("a"^2 + "b"^2) + ("b"^2)/("a"^2 + "b"^2)` = 1

and tan t = `"a"/"b"`

∴ t = `tan^-1("a"/"b")`

∴ y = sin–1(sin t cos x – cos t sin x)

= sin–1[sin(t – x)]

= t – x

= `tan^-1("a"/"b") - x`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x)[tan^-1("a"/"b") - x]`

= 0 – 1

= –1

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अध्याय 2.1: Differentiation - Long Answers III

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