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 (integer)
40
Solution
ln(x + y) = 4xy (At x = 0, y = 1)<br><br>x + y = e<sup>4xy</sup><br><br>$$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$$<br><br>At x = 0 <br><br>${{dy} \over {dx}} = 3$<br><br>$${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$$<br><br>At x = 0, ${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$<br><br>$\Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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