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प्रश्न
`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`
उत्तर
Let y = `tan^-1 [(3"a"^2x - x^3)/("a"^3 - 3"a"x^2)]`
Put x = a tan θ
∴ θ = `tan^-1 x/"a"`
y = `tan^-1 [(3"a"^2 * "a"tantheta - "a"^3 tan^3 theta)/("a"^3 - 3"a"*"a"^2 tan^2theta)]`
⇒ y = `tan^-1 [(3"a"^2 tantheta - "a"^3 tan^3theta)/("a"^3 - 3"a"^3 tan^2theta)]`
⇒ y = `tan^-1 [(3tan theta - tan^2ttheta)/(1 - 3tan^2 theta)]`
⇒ y = `tan^-1 [tan 3theta)]` .......`[because tan 3theta = (3tantheta - tan^2theta)/(1 - 3tan^2theta)]`
⇒ y = 3θ
⇒ y = `3tan^-1 x/"a"`
Differentiating both sides w.r.t. x
`"dy"/"dx" = 3*"d"/"dx" (tan^-1 x/"a")`
= `3* 1/(1 + x^2/"a"^2) * "d"/"dx" * (x/"a")`
= `3 * "a"^2/("a"^2 + x^2) * 1/"a"`
= `(3"a")/("a"^2 + x^2)`
Hence, `"dy"/"dx" = (3"a")/("a"^2 + x^2)`.
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