Rewrite the function using fractional exponents.

To find (dy)/(dx) for y = sqrt(x+x+x) and evaluate it at x=1, we will use the chain rule multiple times. Step 1: Rewrite the function using fractional exponents. y = (x + (x + x^1/2)^1/2)^1/2 Step 2: Differentiate y with respect to x using the chain rule. Let u = x + (x + x^1/2)^1/2. Then y = u^1/2. (dy)/(dx) = (1)/(2) u^-1/2 · (du)/(dx) = (1)/(2sqrt(x+x+x)) · (d)/(dx)(x + (x + x^1/2)^1/2) Now, we need to find (d)/(dx)(x + (x + x^1/2)^1/2): (d)/(dx)(x + (x + x^1/2)^1/2) = 1 + (d)/(dx)((x + x^1/2)^1/2) Let v = x + x^1/2. Then (x + x^1/2)^1/2 = v^1/2. (d)/(dx)(v^1/2) = (1)/(2) v^-1/2 · (dv)/(dx) = (1)/(2sqrt(x+x^1/2)) · (d)/(dx)(x + x^1/2) Now, we need to find (d)/(dx)(x + x^1/2): (d)/(dx)(x + x^1/2) = 1 + (1)/(2)x^-1/2 = 1 + (1)/(2sqrt(x)) Substitute back: (d)/(dx)(sqrt(x+x)) = (1)/(2sqrt(x+x)) · (1 + (1)/(2sqrt(x))) Substitute this back into the expression for (du)/(dx): (du)/(dx) = 1 + (1)/(2sqrt(x+x)) · (1 + (1)/(2sqrt(x))) Finally, substitute (du)/(dx) back into the expression for (dy)/(dx): (dy)/(dx) = (1)/(2sqrt(x+x+x)) (1