 Hi and welcome to the session. Let's work out the following question. The question says, differentiate square root cos x with respect to x from first principle. Let's start with the solution to this question. First of all let y be equal to square root of cos x. Then y plus del y will be equal to square root of cos x plus del x This implies that del y is equal to square root of cos x plus del x minus square root of cos x. Now rationalizing the right hand side we get del y is equal to square root of cos x plus del x minus square root of cos x multiplied by square root of cos x plus del x plus square root of cos x divided by square root cos x plus del x plus square root of cos x. Now here putting the formula that a minus b into a plus b is equal to a square minus b square we have cos of x plus del x minus cos x divided by square root of cos x plus del x plus square root of cos x. Now this is equal to minus 2 sin x plus del x by 2 into sin del x by 2 divided by square root of cos x plus del x plus square root of cos x. Now dividing both the sides y del x we get del y by del x is equal to minus 2 sin x plus del x by 2 into sin del x by 2 divided by del x by 2 multiplied by 1 by 2 multiplied by 1 upon square root of cos x plus del x plus square root of cos x. Now limit del x approaching to 0 del y by del x is equal to limit del x approaching to 0 minus sin x plus del x by 2 multiplied by sin del x by 2 divided by del x by 2 multiplied by 1 upon square root of cos x plus del x plus square root of cos x. This implies that dy by dx is equal to if you put del x to be equal to 0 we get minus sin x. This becomes 1 into 1 upon twice of square root cos x therefore dy by dx is equal to minus sin x divided by 2 into square root of cos x. So our answer to this question is minus sin x divided by 2 into square root of cos x. I hope that you understood the solution and enjoyed the session. Have a good day.