y(t) =√[25² - x(t)²]
y'(t) = (1/2){1/√[25² - x(t)²]}*[-2x(t)]*x'(t)
= -x(t)x'(t)/√[25² - x(t)²]
When x = 9,x'(t) = 3,y'(t) = -9*3/(25² - 9²) = -27/12 = -9/4
Apparently the 2nd question is incomplete.
y(t) =√[25² - x(t)²]
y'(t) = (1/2){1/√[25² - x(t)²]}*[-2x(t)]*x'(t)
= -x(t)x'(t)/√[25² - x(t)²]
When x = 9,x'(t) = 3,y'(t) = -9*3/(25² - 9²) = -27/12 = -9/4
Apparently the 2nd question is incomplete.