How To Answer This Calculus Question?

Labeling the diagram:
(1)
Its x-coordinate increases at a constant rate of 2 in/sec.
Hence, dx/dt = 2

(2)
dy/dt
= dx²/dt
= 2x dx/dt
= 2x (2)
= 4x

(3)
dd/dt
= [d√(x² + y²)]/dt
= {[d√(x² + y²)]/d(x² + y²)} * (dx² + y²)/dt
= 1/[2√(x² + y²)] * [(2x dx/dt) + 2y (dy/dt)]
= 1/[2√(x² + y²)] * {[2x (2)] + 2y (4x)]
= (2x + 4xy)/√(x² + y²)

====
When x = 5:
y = 5²
Hence, y = 25

When x = 5, dd/dt
= (2*5 + 4*5*25)/√(5² + 25²)
= 510/√650
= 510/(5√26)
= 102/√26
= 102(√26)/26
= 51(√26}/13

The rate of change of the distance from the origin when x = 5 is 51(√26}/13 in/sec.

d^2 = (x^2 - 0)^2 + (x - 0)^2
d^2 = x^4 + x^2
2d * dd/dt = 4x^3 * dx/dt + 2x * dx/dt
d * dd/dt = 2x^3 * dx/dt + x * dx/dt
d * dd/dt = (2x^3 + x) * dx/dt
d * dd/dt = x * (2x^2 + 1) * dx/dt
sqrt(x^4 + x^2) * dd/dt = x * (2x^2 + 1) * dx/dt
x * sqrt(x^2 + 1) * dd/dt = x * (2x^2 + 1) * dx/dt
dd/dt = ((2x^2 + 1) / sqrt(x^2 + 1)) * dx/dt

y = x^2
dy/dt = 2x * dx/dt

x = 5
dx/dt = 2

dd/dt = ((2 * 5^2 + 1) / sqrt(5^2 + 1)) * 2
dd/dt = (51 / sqrt(26)) * 2
dd/dt = 102 * sqrt(26) / 26
dd/dt = 51 * sqrt(26) / 13