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प्रश्न
`sec^-1 (1/(4x^3 - 3x)), 0 < x < 1/sqrt(2)`
उत्तर
Let y = `sec^-1 (1/(4x^3 - 3x))`
Put x = cos θ
∴ θ = `cos^-1x`
y = `sec^-1 (1/(4cos^3theta - 3 cos theta))`
⇒ y = `sec^-1 (1/(cos 3theta))` .....[∵ cos 3θ = 4 cos3θ – 3 cos θ]
⇒ y = `sec^-1 (sec 3theta)`
⇒ y = 3θ
y = `3cos^-1x`
Differentiating both sides w.r.t. x
`"dy"/"d" = 3 * "d"/"dx" cos^-1x`
= `3((-1)/sqrt(1 - x^2))`
= `(-3)/sqrt(1 - x^2)`
Hence, `"dy"/"dx" = (-3)/sqrt(1 - x^2)`.
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