JEE MAIN - Mathematics (2013 (Offline) - No. 9)
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
$${1 \over {\sqrt 2 }}$$
$${1 \over 2}$$
$$1$$
$$\sqrt 2 $$
Explanation
Let $$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$$
and $${\tan ^{ - 1}}\,\,x = \theta .$$
$$ \Rightarrow x = \tan \theta $$
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Thus, we have $$y = \sec \,\theta $$
$$ \Rightarrow y = \sqrt {1 + {x^2}} $$
$$\left( {\,\,} \right.$$ As $$\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $$ $$\left. {\,\,} \right)$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$$
At $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$
and $${\tan ^{ - 1}}\,\,x = \theta .$$
$$ \Rightarrow x = \tan \theta $$
Thus, we have $$y = \sec \,\theta $$
$$ \Rightarrow y = \sqrt {1 + {x^2}} $$
$$\left( {\,\,} \right.$$ As $$\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $$ $$\left. {\,\,} \right)$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$$
At $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$
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