JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 18)
If y = y(x) is an implicit function of x such that loge(x + y) = 4xy, then $${{{d^2}y} \over {d{x^2}}}$$ at x = 0 is equal to ___________.
Answer
40
Explanation
ln(x + y) = 4xy (At x = 0, y = 1)
x + y = e4xy
$$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$$
At x = 0
$${{dy} \over {dx}} = 3$$
$${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$$
At x = 0, $${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$$
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$$
x + y = e4xy
$$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$$
At x = 0
$${{dy} \over {dx}} = 3$$
$${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$$
At x = 0, $${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$$
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$$
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