Question

If y = f$\left( \frac{2x - 1}{x^{2} + 1} \right)$ and f ′(x) = sin x², then $\frac{dy}{dx}$ is equal to

• A. cos x². f ′(x)
• B. – cos x². f (x)
• C. $\frac{2(1 + x - x^{2})}{(x^{2} + 1)^{2}}.{\sin\left( \frac{2x - 1}{x^{2} + 1} \right)}^{2}$
• D. None of these

Answer 1

Answer: (C)

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

Answer 2

y = f(2x – 1/ x² + 1)

$\frac{dy}{dx}$ = f ′$\left( \frac{2x - 1}{x^{2} + 1} \right).\frac{(x^{2} + 1).2 - (2x - 1).2x}{(x^{2} + 1)^{2}}$

= sin$\left( \frac{2x - 1}{x^{2} + 1} \right)^{2}.\frac{2x^{2} + 2 - 4x^{2} + 2x}{(x^{2} + 1)^{2}}$