If sec((x2-2x/x2+1))=y, then (dy/dx) is equal to

Question

If sec((x2-2x/x2+1))=y, then (dy/dx) is equal to (A) (y2/x2) (B) (2y√y2-1(x2+x-1)/(x2+1)2) (C) ((x2+x-1)/y√y2-1) (D) (x2-y2/x2+y2)

Tardigrade Admin · Accepted Answer

y=sec((x2-2x/x2+1)) ⇒ (dy/dx)=sec((x2-2x/x2+1))tan((x2-2x/x2+1)). ((x2+1)(2x-2)-(x2-2x)(2x)/(x2+1)2) =sec((x2-2x/x2+1))tan((x2-2x/x2+1))⋅(2x2+2x-2/(x2+1)2) ⇒ (dy/dx)=(2y√y2-1(x2+x-1)/(x2+1)2)

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

$\frac{y ^{2}}{x ^{2}}$

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

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

$\frac{x ^{2} - y ^{2}}{x ^{2} + y ^{2}}$

Q. If sec(x2+1x2−2x​)=y, then dxdy​ is equal to

x2y2​

(x2+1)22yy2−1​(x2+x−1)​

yy2−1​(x2+x−1)​

x2+y2x2−y2​

y=sec(x2+1x2−2x​) ⇒dxdy​=sec(x2+1x2−2x​)tan(x2+1x2−2x​). {(x2+1)2(x2+1)(2x−2)−(x2−2x)(2x)​} =sec(x2+1x2−2x​)tan(x2+1x2−2x​)⋅(x2+1)22x2+2x−2​ ⇒dxdy​=(x2+1)22yy2−1​(x2+x−1)​

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

$\frac{y ^{2}}{x ^{2}}$

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

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

$\frac{x ^{2} - y ^{2}}{x ^{2} + y ^{2}}$