Find the temperature point

The temperature at a point (x, y) is T(x, y), measured in degrees Celsius. A bug crawls so that its position after t seconds is given by the following function, where x and y are measured in centimeters.
x = sqrt(5 + t), y = 2 + 1/4 t
The temperature function satisfies Tx(3, 3) = 4 and Ty(3, 3) = 3. How fast is the temperature rising on the bug's path after 4 seconds? (Round your answer to two decimal places.)
____________°C/s

Please give a detailed response. Thank you!

x = √(5+t)
dx/dt = 1/(2√5+t))

y = 2 + 1/4 t
dy/dt = 1/4

When t = 4
x = √9 = 3 . . . . . dx/dt = 1/(2√9) = 1/6
y = 2+1 = 3 . . . . dy/dt = 1/4

dT/dt = Tx(x,y) dx/dt + Ty(x,y) dy/dt
........ = Tx(3,3) * 1/6 + Ty(3,3) * 1/4
........ = 4 * 1/6 + 3 * 1/4
........ = 2/3 + 3/4
........ = 17/12
........ = 1.42

After 4 seconds, temperature is rising at a rate of 1.42°C/s