x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, so we have the product of two functions of x. You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y. Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0.