Let y = - 1 x dx = (1 + x 2 ) therefore, y d x 2 = ) 0 - 1 (2x) (1 + x 2 ) 2 = (1 + x 2 ) 2

Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.7 ,Q.8,Page 183

Question

${\tan ^{ - 1}}x$

Step-by-step Solution

Let $y = {\tan ^{ - 1}}x$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{1}{{(1 + {x^2})}}$

therefore, $\cfrac{{{d^2}y}}{{d{x^2}}} = \cfrac{{(1 + {x^2}) \cdot 0 - 1 \cdot (2x)}}{{{{(1 + {x^2})}^2}}} = \cfrac{{ - 2x}}{{{{(1 + {x^2})}^2}}}$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Continuity and Differentiability. Curated by Sachin Sharma. Free for all students.