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प्रश्न
Find `"dy"/"dx"` ; if y = cos-1 `("2x" sqrt (1 - "x"^2))`
उत्तर
y = cos-1 `("2x" sqrt (1 - "x"^2))`
Put x = sin θ
∴ θ = sin -1x
∴ Y = cos-1 `(2 "sin" theta sqrt (1 - "x"^2))`
= cos -1 (2 sin θ . cos θ)
= cos-1 `["cos" (pi/2 - 2 theta)]`
`= pi/2 - 2 "sin"^(-1) "x"`
∴ y = - 2 sin-1 x
Differentiating w.r.t.x
`"dy"/"dx" = 0 - 2 "x" 1/sqrt (1 - "x"^2) = (-2)/sqrt (1 - "x"^2)`
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