Question

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

• A. $\frac{y^{2}}{x^{2}}$
• B. $\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}$
• C. $\frac{\left(x^{2}+x-1\right)}{y\sqrt{y^{2}-1}}$
• D. $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

Answer 1

Answer: (B)

$y=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)$
$\Rightarrow \frac{dy}{dx}=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right)$.
$\left\{\frac{\left(x^{2}+1\right)\left(2x-2\right)-\left(x^{2}-2x\right)\left(2x\right)}{\left(x^{2}+1\right)^{2}}\right\}$
$=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right)\cdot\frac{2x^{2}+2x-2}{\left(x^{2}+1\right)^{2}}$
$\Rightarrow \frac{dy}{dx}=\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}$