y = x^(1/√x) = x^(x-¹/²)

Take ln on both sides and perform implicit differentiation.


ln y = ln x^(x-¹/²)

ln y = x-¹/² ln x

Differentiate both sides with respect to x,

1/y dy/dx = -½x-³/² ln x + x-¹/² (1/x)

1/x^(1/√x) dy/dx = (1/√x)(1/x) - lnx / 2x³/²

dy/dx / x^(1/√x) = 1/x³/² - lnx/2 (1/x³/²)

dy/dx / x^(1/√x) = 1/x³/² (1 - lnx/2)

dy/dx = x^(1/√x)/x³/² (1 - lnx/2)

Area of triangle OPQ, A

= ½ × base × height

= ½ × PQ × perpendicular distance from O to PQ

= ½ × y-coordinate of Q × OP

= ½ × x^(1/√x) × x

= ½x (x^(1/√x))

(Since P is on the x-axis, its y-coordinate is 0. The length of OP is basically the x-coordinate of P, but in terms of units. OP is a horizontal line. Recall x > 0 so there's no need to take any modulus/absolute value since x is positive already.

Triangle OPQ is a right-angled triangle and PQ is a vertical line, since P and Q have the same x-coordinate. PQ's length is basically the same as its y-coordinate, but in terms of units.

There is also no need to take modulus since y = x^(1/√x) . x > 0 so this term's base and index is always positive, which means y is always > 0 as well.

Notice that x can never = 0 as 1/√x would be undefined due to division by 0. x also ≮ 0 since the square root would be non-real)


Since P moves along the x-axis from x = 10 to x = 1 at a constant rate of 1 unit per second, the rate of decrease of x is 1 unit per second.

So dx/dt = -1

Integrating both sides with respect to t,
x = -t + c, c is a constant
When t = 0 , x = 10
10 = 0 + c
10 = c

So x = -t + 10
x = 10 - t

(Actually there is no need to do integration and finding the constant c as it should be quite obvious that if x is decreasing at 1 unit/s, then x = 10 - 1/second × t seconds → x = 10 - t)

Now we already have the result of dy/dx in the first part.

So we differentiate A using product rule, instead of combining the x terms together, and then having to take ln on both sides and doing implicit differentiation again.

A = ½x (x^(1/√x))

dA/dx = ½ [ (1) x^(1/√x) + x (x^(1/√x)/x³/² (1 - lnx/2)) ]

dA/dx = ½ [ x^(1/√x) + x^(1/√x)/√x (1 - lnx/2)]

dA/dx = ½x^(1/√x) [1 + (1 - lnx/2)/√x]


When t = 6, x = 10 - 6 = 4

Then,

dA/dx = ½(4^(1/√4)) [1 + (1 - ln4 /2)/√4]

dA/dx = ½(4¹/²) [1 + (1 - ln2² / 2)/2]

dA/dx = ½(2) [1 + ½(1 - 2ln2 / 2)]

dA/dx = 1 + ½ - ½ln2

= 3/2 - ½ln2


Now the area is measured/calculated in terms of unit² and x is in terms of unit.

So dA/dx , which is the change in area with respect to change in length (or base of triangle), would be in terms of unit²/unit = unit


Then, rate of change of A is denoted as dA/dt

dA/dt = dA/dx × dx/dt

= (3/2 - ½ln2) × (-1)

= ½ln2 - 3/2


dx/dt, which is the rate of change of the x-coordinate (rate means with respect time) has units of unit/s or unit s-¹


So dA/dt would have units of :

unit × unit s-¹ → unit² s-¹


So rate of change of A, dA/dt = (½ln2 - 3/2) unit² s-¹