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Mathematics Question on Continuity and differentiability

If sec(x2−2xx2+1)=ysec\left(\frac{x^{2}-2x}{x^{2}+1}\right)=y, then dydx\frac{dy}{dx} is equal to

A

y2x2\frac{y^{2}}{x^{2}}

B

2yy2−1(x2+x−1)(x2+1)2\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}

C

(x2+x−1)yy2−1\frac{\left(x^{2}+x-1\right)}{y\sqrt{y^{2}-1}}

D

x2−y2x2+y2\frac{x^{2}-y^{2}}{x^{2}+y^{2}}

Answer

2yy2−1(x2+x−1)(x2+1)2\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}

Explanation

Solution

y=sec(x2−2xx2+1)y=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right) ⇒dydx=sec(x2−2xx2+1)tan(x2−2xx2+1)\Rightarrow \frac{dy}{dx}=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right). \left\\{\frac{\left(x^{2}+1\right)\left(2x-2\right)-\left(x^{2}-2x\right)\left(2x\right)}{\left(x^{2}+1\right)^{2}}\right\\} =sec(x2−2xx2+1)tan(x2−2xx2+1)⋅2x2+2x−2(x2+1)2=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right)\cdot\frac{2x^{2}+2x-2}{\left(x^{2}+1\right)^{2}} ⇒dydx=2yy2−1(x2+x−1)(x2+1)2\Rightarrow \frac{dy}{dx}=\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}