 Hello friends and how are you all doing today? It says, if y is equal to tan inverse bracket 1 plus x square in the square root plus under root 1 minus x square upon under root 1 plus x square minus under root 1 minus x square, fine dy by dx. So here we are given y as tan inverse bracket under root 1 plus x square plus under root 1 minus x square upon under root 1 plus x square minus under root 1 minus x square. We need to find dy by dx, fine dy by dx, right? So here we can put x square as cos 2 right on doing so we have y is equal to tan inverse under root 1 plus 2 theta plus under root 1 minus cos 2 theta upon under root 1 plus cos 2 theta minus under root 1 minus cos 2. We know that 1 plus cos 2 theta is equal to 2 cos square theta, right? And 1 minus cos 2 theta is 2 sine square theta, right? On using it we have y is equal to tan inverse bracket under root 2 cos square theta 2 sine square theta. So we have tan inverse under root cos theta upon under root cos theta theta minus in the denominator we have root 2 sine. On taking out root 2 common from numerator and denominator and on cancelling them we have y is equal to tan inverse cos theta plus sine theta cos theta minus sine theta. Now on dividing numerator and denominator by cos theta we have 1 plus sine theta upon cos theta will give us tan theta y is equal to tan inverse tan theta upon 1 minus tan theta we have y is equal to pi by 4 plus 1 by 2 cos inverse y is equal to plus 1 by 2 cos inverse. On differentiating both the sides with respect to x we have dy by dx equal to dy dx of pi by 4 1 by 2 cos inverse x. So we have dy by dx equal to derivative of a constant is 0 plus this will give us 1 by 2 minus 1 upon under root 1 minus x square is r x itself so it will be x raised to the power 4 into derivative of x square which is we have dy by dx equal to minus x upon under root 1 minus x raised to the power 4 and this is a required answer to this question. So hope you understood it well and enjoyed it too have a nice day ahead.