Question

If y = $\frac{1}{x}$, then the value of $\frac{dy}{\sqrt{1 + y^{4}}} + \frac{dx}{\sqrt{1 + x^{4}}} + 3$ is equal to-

• A. 0
• B. 3
• C. 4
• D. –3

Answer 1

Answer: (B)

3

Answer 2

y = $\frac{1}{x}$ ⇒ $\frac{dy}{dx}$ = $- \frac{1}{x^{2}}$

⇒ x²dy + dx = 0

⇒ $\frac{x^{2}}{\sqrt{1 + x^{4}}}dy + \frac{dx}{\sqrt{1 + x^{4}}} = 0$

⇒ $\frac{dy}{\sqrt{\frac{1}{x^{4}} + 1}} + \frac{dx}{\sqrt{1 + x^{4}}}$ = 0

⇒ $\frac{dy}{\sqrt{1 + y^{4}}} + \frac{dx}{\sqrt{1 + x^{4}}}$ = 0

⇒ $\frac{dy}{\sqrt{1 + y^{4}}} + \frac{dx}{\sqrt{1 + x^{4}}} + 3$ = 3