1
JEE Main 2017 (Offline)
+4
-1
If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals
A
$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$
B
$${{{3x} \over {1 - 9{x^3}}}}$$
C
$${{3 \over {1 + 9{x^3}}}}$$
D
$${{9 \over {1 + 9{x^3}}}}$$
2
JEE Main 2014 (Offline)
+4
-1
If $$g$$ is the inverse of a function $$f$$ and $$f'\left( x \right) = {1 \over {1 + {x^5}}},$$ then $$g'\left( x \right)$$ is equal to:
A
$${1 \over {1 + {{\left\{ {g\left( x \right)} \right\}}^5}}}$$
B
$$1 + {\left\{ {g\left( x \right)} \right\}^5}$$
C
$$1 + {x^5}$$
D
$$5{x^4}$$
3
JEE Main 2013 (Offline)
+4
-1
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}$$
C
$$1$$
D
$$\sqrt 2$$
4
AIEEE 2011
+4
-1
$${{{d^2}x} \over {d{y^2}}}$$ equals:
A
$$- {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
B
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
C
$$- \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
D
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
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