If cos (f(x))=(1-x2/1+x2) and tan (g(x))=(3 x-x3/1-3 x2) then (d f/d g)=

Question

If cos (f(x))=(1-x2/1+x2) and tan (g(x))=(3 x-x3/1-3 x2) then (d f/d g)= (A) (3/2) (B) (1+x2+2 x3/(1-3 x2)2) (C) (2/3) (D) (x2+x3/(1+x2)(1-3 x2))

Tardigrade Admin · Accepted Answer

Given, cos (f(x))=(1-x2/1+x2) Put x= tan θ cos (f(x))=(1- tan 2 θ/1+ tan 2 θ) cos (f(x))= cos 2 θ f(x)=2 θ=2 tan -1 x tan (g(x))=(3 x-x3/1-3 x2) Put x= tan θ tan (g(x)) =(3 tan θ- tan 3 θ/1-3 tan 2 θ) tan (g(x)) = tan 3 θ ∴ g(x) =3 θ=3 tan -1 x ∴ (f'(x)/g'(x)) =(2/3) ∴ (d f/d g) =(2/3)

Q. If $cos (f (x)) = \frac{1 - x ^{2}}{1 + x ^{2}}$ and $tan (g (x)) = \frac{3 x - x ^{3}}{1 - 3 x ^{2}}$ then $\frac{df}{d g} =$

$\frac{3}{2}$

$\frac{1 + x ^{2} + 2 x ^{3}}{( 1 - 3 x ^{2} ) ^{2}}$

$\frac{2}{3}$

$\frac{x ^{2} + x ^{3}}{( 1 + x ^{2} ) ( 1 - 3 x ^{2} )}$

Q. If cos(f(x))=1+x21−x2​ and tan(g(x))=1−3x23x−x3​ then dgdf​=

23​

(1−3x2)21+x2+2x3​

32​

(1+x2)(1−3x2)x2+x3​

Given, cos(f(x))=1+x21−x2​ Put x=tanθ cos(f(x))=1+tan2θ1−tan2θ​ cos(f(x))=cos2θ f(x)=2θ=2tan−1x tan(g(x))=1−3x23x−x3​ Put x=tanθ tan(g(x))=1−3tan2θ3tanθ−tan3θ​ tan(g(x))=tan3θ ∴g(x)=3θ=3tan−1x ∴g′(x)f′(x)​=32​ ∴dgdf​=32​

Q. If $cos (f (x)) = \frac{1 - x ^{2}}{1 + x ^{2}}$ and $tan (g (x)) = \frac{3 x - x ^{3}}{1 - 3 x ^{2}}$ then $\frac{df}{d g} =$

$\frac{3}{2}$

$\frac{1 + x ^{2} + 2 x ^{3}}{( 1 - 3 x ^{2} ) ^{2}}$

$\frac{2}{3}$

$\frac{x ^{2} + x ^{3}}{( 1 + x ^{2} ) ( 1 - 3 x ^{2} )}$