d dx ( e^ 1 - x^2 . x ) - Vidyadip

MCQ

$\frac{d}{{dx}}\left( {{e^{\sqrt {1 - {x^2}} }}.\tan x} \right)$

A
${e^{\sqrt {1 - {x^2}} }}\left[ {{{\sec }^2}x + \frac{{x\tan x}}{{\sqrt {1 - {x^2}} }}} \right]$
✓
${e^{\sqrt {1 - {x^2}} }}\left[ {{{\sec }^2}x - \frac{{x\tan x}}{{\sqrt {1 - {x^2}} }}} \right]$
C
${e^{\sqrt {1 - {x^2}} }}\left[ {{{\sec }^2}x + \frac{{\tan x}}{{\sqrt {1 - {x^2}} }}} \right]$
D
None of these

✓

Answer

Correct option: B.

${e^{\sqrt {1 - {x^2}} }}\left[ {{{\sec }^2}x - \frac{{x\tan x}}{{\sqrt {1 - {x^2}} }}} \right]$

b
(b) $\frac{d}{{dx}}\left[ {{e^{\sqrt {1 - {x^2}} }}\tan x} \right]$
$ = {e^{\sqrt {1 - {x^2}} }}\frac{1}{{2\sqrt {1 - {x^2}} }}( - 2x)\tan x + {e^{\sqrt {1 - {x^2}} }}{\sec ^2}x$
$ = {e^{\sqrt {1 - {x^2}} }}\left[ {{{\sec }^2}x - \frac{{x\tan x}}{{\sqrt {1 - {x^2}} }}} \right]$.

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