If y = f((2x-1/x2 + 1)) and f'(x) = sin x2 , then (dy/dx) is equal to

Question

If y = f((2x-1/x2 + 1)) and f'(x) = sin x2 , then (dy/dx) is equal to (A)  cos x2 .f'(x) (B)  - cos x2 .f'(x) (C) (2(1+x-x2)/(x2+1)2) sin ((2x-1/x2 + 1))2 (D) none of these

Tardigrade Admin · Accepted Answer

(dy/dx) = f'(z) (dz/dx) where z = (2x-1/x2 + 1) = sin z2 . ((x2 + 1)2 - (2x -1) 2x/(x2 + 1)2) = sin((2x-1/x2+ 1))2 . (2+2x-2x2/(x2 + 1 )2)

Q. If $y = f (\frac{2 x - 1}{x ^{2} + 1})$ and $f^{'} (x) = sin x^{2}$ , then $\frac{d y}{d x}$ is equal to

$cos x^{2} . f^{'} (x)$

$- cos x^{2} . f^{'} (x)$

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

none of these

$\frac{d y}{d x} = f^{'} (z) \frac{d z}{d x} w h ere z = \frac{2 x - 1}{x ^{2} + 1}$
$= sin z^{2} . \frac{( x ^{2} + 1 ) 2 - ( 2 x - 1 ) 2 x}{( x ^{2} + 1 ) ^{2}}$
$= sin (\frac{2 x - 1}{x ^{2} + 1})^{2} . \frac{2 + 2 x - 2 x ^{2}}{( x ^{2} + 1 ) ^{2}}$

Q. If y=f(x2+12x−1​) and f′(x)=sinx2, then dxdy​ is equal to

cosx2.f′(x)

−cosx2.f′(x)

(x2+1)22(1+x−x2)​sin(x2+12x−1​)2