MCQ
જો $y = {e^{{{\tan }^{ - 1}}x}}$, તો $(1 + {x^2}){{{d^2}y} \over {d{x^2}}} = $
- ✓$(1 - 2x){{dy} \over {dx}}$
- B$ - 2x{{dy} \over {dx}}$
- C$ - x{{dy} \over {dx}}$
- D$0$
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{(1 + {x^2}).\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{(1 + {x^2})}} - {e^{{{\tan }^{ - 1}}x}}(2x)}}{{{{(1 + {x^2})}^2}}}$
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{(1 - 2x){e^{{{\tan }^{ - 1}}x}}}}{{{{(1 + {x^2})}^2}}}$
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}}(1 + {x^2}) = (1 - 2x)\frac{{dy}}{{dx}}$.
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કારણ $(R) : \,$ જો $\overline {{\text{AB}}} \,\, = \,\,\vec a ,\;\,\overline {BC} \,\,\, = \,\,\vec b \,$ તો $\overline {AC} = \,\vec a + \,\,\vec b $ (સરવાળા ત્રિકોણ નિયમ )