The derivative of tan- 1[((3 x2 - 1)/(3 x - x3))] with respect to sin- 1[((x2 - 1)/(x2 + 1))] is

Question

The derivative of tan- 1[((3 x2 - 1)/(3 x - x3))] with respect to sin- 1[((x2 - 1)/(x2 + 1))] is (A) (2/3) (B) (- 2/3) (C) (3/2) (D) -(3/23)

Tardigrade Admin · Accepted Answer

y=(tan)- 1((3 x2 - 1/3 x - x3)) =(π /2)-tan1((3 x - x3/3 x2 - 1))=(π /2)+3(tan)1x (d y/d x)=(3/1 + x2) u=(sin)- 1((x2 - 1/x2 + 1))=(π /2)-(cos)- 1((x2 - 1/x2 + 1)) =-(π /2)+2tan- 1x (d u/d x)=(2/1 + x2) (d y/d x)=((d y/d u)/(d u)d x)=(3/2)

Q. The derivative of $t a n^{- 1} [\frac{( 3 x ^{2} - 1 )}{( 3 x - x ^{3} )}]$ with respect to $s i n^{- 1} [\frac{( x ^{2} - 1 )}{( x ^{2} + 1 )}]$ is

$\frac{2}{3}$

$\frac{- 2}{3}$

$\frac{3}{2}$

$- \frac{3}{23}$

Q. The derivative of tan−1[(3x−x3)(3x2−1)​] with respect to sin−1[(x2+1)(x2−1)​] is

32​

3−2​

23​

−233​

y=(tan)−1(3x−x33x2−1​) =2π​−tan1(3x2−13x−x3​)=2π​+3(tan)1x dxdy​=1+x23​ u=(sin)−1(x2+1x2−1​)=2π​−(cos)−1(x2+1x2−1​) =−2π​+2tan−1x dxdu​=1+x22​ dxdy​=dxdu​dudy​​=23​

Q. The derivative of $t a n^{- 1} [\frac{( 3 x ^{2} - 1 )}{( 3 x - x ^{3} )}]$ with respect to $s i n^{- 1} [\frac{( x ^{2} - 1 )}{( x ^{2} + 1 )}]$ is

$\frac{2}{3}$

$\frac{- 2}{3}$

$\frac{3}{2}$

$- \frac{3}{23}$