 Hi and welcome to the session, let us discuss following question, question says, if x multiplied by under root of 1 plus y plus y multiplied by under root of 1 plus x is equal to 0 for x greater than minus 1 and less than 1 prove that dy upon dx is equal to minus 1 upon plus x whole square. Let us start the solution now, we are given x multiplied by under root of 1 plus y plus y multiplied by under root of 1 plus x is equal to 0. Now, this equation can be written as x multiplied by under root of 1 plus y is equal to minus y multiplied by square root of 1 plus x. Here we have subtracted this term from both the sides. Now, squaring both the sides, we get x square multiplied by 1 plus y is equal to y square multiplied by 1 plus x. We know square of this term is equal to this and square of this term is equal to this. Now, this can be further written as x square plus x square y is equal to y square plus y square x. We have multiplied x square by this bracket and y square by this bracket. Now, subtracting y square from both the sides, we get x square minus y square plus x square y is equal to y square x. Now, subtracting x square y from both the sides, we get x square minus y square is equal to y square x minus x square y. Now, here we will apply the formula for a square minus b square. We know a square minus b square is equal to a plus b multiplied by a minus b. So, we can write x square minus y square is equal to x plus y multiplied by x minus y. On right hand side, we will take x y common and we will get y minus x inside the bracket. Now, this can be further written as x plus y multiplied by x minus y is equal to minus x y multiplied by x minus y. Taking minus 1 common from this bracket, we get minus x y multiplied by x minus y. Now, this bracket and this bracket will cancel each other. Now, we get x plus y is equal to minus x y. Now, subtracting y from both the sides, we get x is equal to minus x y minus y. Now, we will take minus y common on right hand side. Now, taking minus y common, we are left with x plus 1 in the bracket. Now, we get x is equal to minus y multiplied by x plus 1. Now, dividing both the sides by x plus 1, we get x upon x plus 1 is equal to minus y or we can simply write it as y is equal to minus of x upon x plus 1. Here, we have multiplied both the sides by minus 1 to get the value of y. Now, let us find out derivative of y with respect to x. Now, differentiating both the sides with respect to x, we get dy upon dx. We will calculate derivative of this term by quotient rule. Now, this would be equal to x plus 1 multiplied by derivative of minus x minus minus x multiplied by derivative of x plus 1 upon x plus 1 whole square. Now, this implies dy upon dx is equal to x plus 1 multiplied by minus 1. We know derivative of minus x is equal to minus 1 minus minus x multiplied by 1. We know derivative of x plus 1 is equal to 1 upon x plus 1 whole square. Now, simplifying, we get dy upon dx is equal to minus x minus 1 plus x upon x plus 1 whole square. Now, x and plus x will cancel each other and we get dy upon dx is equal to minus 1 upon x plus 1 whole square. So, we get dy upon dx is equal to minus 1 upon x plus 1 whole square. This is our required answer. This completes the session. Hope you understood the session. Take care and have a nice day.