If y x + x + x^2 y = 0, then dy dx = - Vidyadip

MCQ

If $y\sec x + \tan x + {x^2}y = 0$, then ${{dy} \over {dx}} =$

A
${{2xy + {{\sec }^2}x + y\sec x\tan x} \over {{x^2} + \sec x}}$
B
$ - {{2xy + {{\sec }^2}x + \sec x\tan x} \over {{x^2} + \sec x}}$
✓
$ - {{2xy + {{\sec }^2}x + y\sec x\tan x} \over {{x^2} + \sec x}}$
D
None of these

✓

Answer

Correct option: C.

$ - {{2xy + {{\sec }^2}x + y\sec x\tan x} \over {{x^2} + \sec x}}$

c
(c) $y\sec x + \tan x + {x^2}y = 0$

$ \Rightarrow \sec x\frac{{dy}}{{dx}} + y\sec x\tan x + {\sec ^2}x + 2xy + {x^2}\frac{{dy}}{{dx}} = 0$

==> $\frac{{dy}}{{dx}} = - \frac{{2xy + {{\sec }^2}x + y\sec x\tan x}}{{{x^2} + \sec x}}$.

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