 Hi and welcome to the session. Let's work out the following question. The question says differentiate the following with respect to x that is log x plus square root 1 plus x square So let us start with a solution to this question Now we are given Say y is equal to log x plus square root 1 plus x square Now differentiating both the sides with respect to x we get dy by dx is equal to 1 upon x plus square root 1 plus x square into d by dx of the function that is x plus Square root 1 plus x square now this is same as 1 upon x plus square root 1 plus x square now derivative with respect to x of x is 1 So we have 1 plus now derivative with respect to x of square root 1 plus x square is 2x divided by 2 into square root 1 plus x square Therefore we can say that dy by dx is 1 into square root 1 plus x square plus x divided by x plus square root 1 plus x square into square root 1 plus x square and This is equal to square root 1 plus x square because here we see that x plus square root 1 plus x squared gets cancelled from numerator and denominator So our answer to this question is 1 upon square root 1 plus x square So I hope that you understood the solution and enjoyed the session. Have a good day