Question

Tan-1 ( 3a2x-x3 / a3 - 3ax2), a>0; -a/√3 < x < a/√3

Answer 1

Answer:

Step-by-step explanation:

माना  

           $tan^{-1}(\dfrac{x}{a} )=\theta.i.e., x=\tan\theta$

अब  

       $tan^{-1}(\dfrac{3a^2x-x^3}{a^3-3ax^2})\\ \\\\=tan^{-1}( \dfrac{3a^2(a\tan\theta)-(a\tan \theta)^3}{a^3-3a(atan\theta)^2})\\\\\\=tan^{-1}(\dfrac{a^3(3tan\theta-tan^3\theta)}{a^3(1-3tan^2\theta)} )\\\\\\=tan^{-1}(\dfrac{3tan\theta-tan^3\theta}{1-3tan^2\theta})\\ \\\\=tan^{-1}(\tan 3\theta)\\\\=3\theta\\\\\\=3tan^{-1}(x/a)$

Answer 2

y = tan-1(3a2x-x3/a3-3ax2)

Let x2 = atanQ

Q = tan-1(x/a)

y = tan-1(3tanQ-tan3Q/1-3tan2Q)

= tan-1(tan3Q)

= 3Q

= 3tan-1(x/a)

dy/dx=3/a(1+x2/a2)

= 3a/a2+x2