Verified answer

Let Δx = a positive number (change in x)

Let m = the slope of the tangent line to y = √x from x = 4 to x = 4 + Δx.

When x = 4,

we get y = √4 or

y = 2.

referring to the point (4,2).

When x = 4 + Δx,

we get y = √(4 + Δx)

referring to the point (4 + Δx,√(4 + Δx))

Now, we use the slope for mula to find m (slope of the secant line):

m = [√(4 + Δx) - 2]/[(4 + Δx) - 4]

We then get:

m = [√(4 + Δx) - 2]/Δx

Now consider the limit of m as Δx approaches zero. Gradually, the secant line becomes the tangent line at x = 4. This is what we want, so

(letting µ = slope of the tangent line,)

µ = lim Δx→0 [√(4 + Δx) - 2]/Δx

We then simplify as follows

Multiply the function by [√(4 + Δx) + 2]/[√(4 + Δx) + 2] (referring to the conjugate of the numerator)

= lim Δx→0 [√(4 + Δx) - 2]/Δx · [√(4 + Δx) + 2]/[√(4 + Δx) + 2]

Thus we get:

= lim Δx→0 [(4 + Δx) - 4]/Δx[√(4 + Δx) + 2]

then,

= lim Δx→0 Δx/Δx[√(4 + Δx) + 2]

then,

= lim Δx→0 1/[√(4 + Δx) + 2]

From here, we can now substitute Δx = 0.

= 1/[√(4 + 0) + 2]

= 1/[√4 + 2]

= 1/(2 + 2)

= 1/4

Therefore, the slope of the tangent line to y = √x at x = 4 is 1/4.

^_^

y(x) = x^(1/2)

Take the derivative to find a the equation of the tangent lines, which also represents the instantaneous rate of change (slope).

y'(x) = dy/dx = (1/2)x^(-1/2)

Plug in x = 4.

y'(4) = (1/2)*4*(-1/2) = 1/4