MCQ
${\left[ {\frac{d}{{dx}}{{\sec }^{ - 1}}x} \right]_{x = - 3}} = ..........$
- A$\frac{1}{{\sqrt {{x^2} - 1} }}$
- B$ - \frac{1}{{\sqrt {{x^2} - 1} }}$
- ✓$\frac{1}{{6\sqrt 2 }}$
- D$-\frac{1}{{6\sqrt 2 }}$
${\left[ {\frac{d}{{dx}}{{\sec }^{ - 1}}x} \right]_{x = - 3}} = {\left[ {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}} \right]_{x = - 3}}$
$ = \frac{1}{{\left| { - 3} \right|\sqrt {{{\left( { - 3} \right)}^2} - 1} }}$
$=\frac{1}{{3\sqrt {9 - 1} }}$
$ = \frac{1}{{3\sqrt 8 }} = \frac{1}{{6\sqrt 2 }}$
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