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Question

`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`

Sum
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Solution

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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Chapter 5: Continuity And Differentiability - Exercise [Page 110]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 42 | Page 110

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