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प्रश्न
if `y = sin^(-1) (6xsqrt(1-9x^2))`, `1/(3sqrt2) < x < 1/(3sqrt2)` then find `(dy)/(dx)`
उत्तर
`y = sin^(-1) (6x sqrt(1-9x^2)), -1/(3sqrt2) < x < 1/(3sqrt2)`
`=> y = sin^(-1) (2 xx 3x xx sqrt(1-(3x)^2))`
Putting 3x = sinθ, we have
`y = sin^(-1) (2sin theta sqrt(1- sin^2 theta))`
⇒ y = sin−1 (2sinθ cosθ)
⇒ y = sin−1(sin2θ)
⇒ y = 2θ
`[-1/(3sqrt2) < (sin theta)/3 < 1/(3sqrt2) => - 1/sqrt2 < sin theta < 1/sqrt2 => - pi/4 < theta < pi/4 => -pi/4 < 2theta < pi/2]`
`:. y = 2sin^(-1) 3x`
Differentiating both sides w.r.t x, we get
`(dy)/(dx) = 2 xx 1/(sqrt(1-(3x)^2)) xx 3`
`=> (dy)/dx = 6/sqrt(1- 9x^2)`
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