Verified answer

If f(x) = 1/√(2x + 3), then:

f(x + Δx) = 1/√[2(x + Δx) + 3] = 1/√(2x + 2Δx + 3).

So:

[f(x + Δx) - f(x)]/Δx

= [1/√(2x + 2Δx + 3) - 1/√(2x + 3)]/Δx.

Multiplying the numerator and denominator by √[(2x + 3)(2x + 2Δx + 3)] gives:

[1/√(2x + 2Δx + 3) - 1/√(2x + 3)]/Δx

= [√(2x + 3) - √(2x + 2Δx + 3)]/{Δx√[(2x + 3)(2x + 2Δx + 3)]}.

Then, rationalizing the numerator by multiplying the numerator and denominator by √(2x + 3) + √(2x + 2Δx + 3) yields:

[√(2x + 3) - √(2x + 2Δx + 3)]/{Δx√[(2x + 3)(2x + 2Δx + 3)]}

= {[√(2x + 3) + √(2x + 2Δx + 3)] * [√(2x + 3) - √(2x + 2Δx + 3)]}/{Δx√[(2x + 3)(2x + 2Δx + 3)] * [√(2x + 3) + √(2x + 2Δx + 3)]}

= [(2x + 3) - (2x + 2Δx + 3)]/{Δx√[(2x + 3)(2x + 2Δx + 3)] * [√(2x + 3) + √(2x + 2Δx + 3)]}

(via difference of squares)

= (-2Δx)/{Δx√[(2x + 3)(2x + 2Δx + 3)] * [√(2x + 3) + √(2x + 2Δx + 3)]}

= -2/{√[(2x + 3)(2x + 2Δx + 3)] * [√(2x + 3) + √(2x + 2Δx + 3)]}, by canceling Δx.

Then, letting Δx --> 0 gives:

-2/{√[(2x + 3)(2x + 0 + 3)] * [√(2x + 3) + √(2x + 0 + 3)]}

= -2/{√[(2x + 3)(2x + 3)] * [√(2x + 3) + √(2x + 3)]}

= -2/{(2x + 3) * [2√(2x + 3)]}

= -1/(2x + 3)^(3/2).

I hope this helps!