MCQ
$y = \sec ({\tan ^{ - 1}}x)$ નું વિકલ સમીકરણ મેળવો.
- A$(1 + {x^2})\frac{{dy}}{{dx}} = y + x$
- B$(1 + {x^2})\frac{{dy}}{{dx}} = y - x$
- ✓$(1 + {x^2})\frac{{dy}}{{dx}} = xy$
- D$(1 + {x^2})\frac{{dy}}{{dx}} = \frac{x}{y}$
$\frac{{dy}}{{dx}} = \sec ({\tan ^{ - 1}}x)\tan ({\tan ^{ - 1}}x)\,.\,\frac{1}{{1 + {x^2}}}$$ = \frac{{xy}}{{1 + {x^2}}}$
==> $(1 + {x^2})\frac{{dy}}{{dx}} = xy$.
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