 Hi and welcome to the session. Let us discuss the following question which says differentiate the following with respect to x tan inverse square root of 1 plus x minus square root of 1 minus x upon square root of 1 plus x plus square root of 1 minus x. Before moving on to the solution let's see some formulas which we will use to solve this question. First of all we have 1 plus cos 2 theta is equal to 2 cos square theta. Then we have 1 minus cos 2 theta is equal to 2 sin square theta. Third is tan A minus B is equal to tan A minus tan B upon 1 plus tan A into tan B. Next is derivative of cos inverse x is equal to minus 1 upon square root of 1 minus x square. So these formulas will work as the key idea for this question. Now let us proceed with its solution. We need to differentiate tan inverse of square root of 1 plus x minus square root of 1 minus x upon square root of 1 plus x plus square root of 1 minus x. Let us take it as y. Here to differentiate this we will substitute the value of x as cos 2 theta. So let us put x is equal to cos 2 theta. So now y will be equal to tan inverse of square root of 1 plus cos 2 theta minus square root of 1 minus cos 2 theta upon square root of 1 plus cos 2 theta plus square root of 1 minus cos 2 theta. Now by the key idea we know that 1 plus cos 2 theta is 2 cos square theta and 1 minus cos 2 theta is 2 sin square theta. So here this will be equal to tan inverse square root of 2 cos square theta minus square root of 2 sin square theta upon square root of 2 cos square theta plus square root of 2 sin square theta which will be equal to tan inverse root 2 cos theta minus root 2 sin theta upon root 2 cos theta plus root 2 sin theta. Now root 2 will get cancelled from the numerator and denominator and we are left with tan inverse cos theta minus sin theta upon cos theta plus sin theta. Now let us define the numerator and denominator by cos theta. So we will get tan inverse cos theta upon cos theta minus sin theta upon cos theta upon cos theta upon cos theta plus sin theta upon cos theta dividing numerator and denominator by cos theta. So this will be equal to tan inverse cos upon cos theta is 1 minus sin theta upon cos theta is tan theta upon cos theta upon cos theta that is 1 plus sin theta upon cos theta that is tan theta. So we get tan inverse of 1 minus tan theta upon 1 plus tan theta. Now if we have some number x then we can write x as 1 into x. So here let us write tan theta as 1 into tan theta. Now we know that tan pi by 4 is equal to 1. So in this let us replace 1 by tan pi by 4. Now we have tan A minus B is equal to tan A minus tan B upon 1 plus tan A tan B. So here we will replace this one and this one by tan pi by 4 and this will be equal to tan inverse of tan pi by 4 minus tan theta upon 1 plus tan pi by 4 into tan theta. Now if we look at this formula then here A is pi by 4 and B is theta. So this will be equal to tan inverse of tan pi by 4 minus theta that is tan A minus B. Now we know that tan inverse of tan theta is equal to theta. So this will be equal to pi by 4 minus theta. So now we have y is equal to pi by 4 minus theta and we need to find dy by dx that is derivative of y with respect to x. Now here we took x equal to cos 2 theta. So now x is equal to cos 2 theta. So this implies theta will be equal to 1 by 2 cos inverse x. So here let us replace theta by 1 by 2 cos inverse x. Thus y is equal to pi by 4 minus 1 by 2 cos inverse x. Now let us differentiate y with respect to x. So dy by dx will be equal to 0 minus 1 by 2 into differentiation of cos inverse x as the differentiation of a constant is 0. Now we already know from the key idea the differentiation of cos inverse x is minus 1 upon square root of 1 minus x square. So here dy by dx will be equal to minus 1 by 2 into minus 1 upon square root of 1 minus x square. So this will be equal to 1 upon 2 into square root of 1 minus x square. Thus dy by dx is equal to 1 upon 2 into square root of 1 minus x square is the required answer to this question. With this we finish this session. Hope you must have understood the question. Goodbye, take care and have a nice day.