Difference Quotient for Square Root Functions

The difference quotient for sqrt(x) gives you (sqrt(x+h)-sqrt(x))/h. You get stuck because you cannot cancel h from the top. Multiply top and bottom by the conjugate: sqrt(x+h)+sqrt(x). That turns the top into (x+h)-x = h, and the h cancels. The difference quotient square root problem collapses to 1/(sqrt(x+h)+sqrt(x)).

The conjugate method for square roots extends to cube roots. It covers the common errors that keep the cancellation from working.

With polynomials, expanding (x+h)^2 produces terms that contain a factor of h, which you can factor out. With sqrt(x), the top sqrt(x+h)-sqrt(x) has no common factor of h on its own. The conjugate multiplication creates a difference of squares: (a-b)(a+b)=a^2-b^2. Here a = sqrt(x+h), b = sqrt(x), so a^2-b^2 = (x+h)-x = h. That h now sits as a factor in the top, ready to cancel with the h in the bottom.

Without the conjugate, you cannot move forward. The expression stays in its raw form and the limit as h->0 is not reachable.

How The Conjugate Works Step By Step

Start with (sqrt(x+h)-sqrt(x))/h. Multiply top and bottom by sqrt(x+h)+sqrt(x). The top becomes (x+h)-x = h. The bottom becomes h(sqrt(x+h)+sqrt(x)). Cancel the common factor h from top and bottom. The simplified difference quotient is 1/(sqrt(x+h)+sqrt(x)).

The failure mode here is stopping after writing the conjugate pair but not simplifying the top fully. Write the bottom as h(sqrt(x+h)+sqrt(x)) and then cancel.

Let f(x) = sqrt(x). Compute the difference quotient.

f(x+h) = sqrt(x+h). The difference quotient is (sqrt(x+h)-sqrt(x))/h.

Multiply top and bottom by the conjugate sqrt(x+h)+sqrt(x):

((sqrt(x+h)-sqrt(x)) * (sqrt(x+h)+sqrt(x))) / (h * (sqrt(x+h)+sqrt(x)))

Simplify the top using the difference of squares: (x+h)-x = h.

The expression becomes h / (h * (sqrt(x+h)+sqrt(x))). Cancel h to get 1/(sqrt(x+h)+sqrt(x)).

Now take the limit as h->0: 1/(sqrt(x)+sqrt(x)) = 1/(2sqrt(x)). That is the derivative of sqrt(x).

The mistake students make: dividing the original top by h without rationalizing. That gives (sqrt(x+h)-sqrt(x))/h with no cancellation possible.

Let f(x) = sqrt(2x+5). The shifted input is f(x+h) = sqrt(2(x+h)+5) = sqrt(2x+2h+5).

The difference quotient: (sqrt(2x+2h+5) - sqrt(2x+5)) / h.

Multiply top and bottom by the conjugate: sqrt(2x+2h+5) + sqrt(2x+5).

Top becomes (2x+2h+5) - (2x+5) = 2h.

The expression is 2h / (h * (sqrt(2x+2h+5) + sqrt(2x+5))). Cancel h to get 2/(sqrt(2x+2h+5) + sqrt(2x+5)).

Limit as h->0: 2/(sqrt(2x+5)+sqrt(2x+5)) = 2/(2 sqrt(2x+5)) = 1/sqrt(2x+5). The derivative of sqrt(2x+5) is 1/sqrt(2x+5), which matches the chain rule.

The common error here is incorrectly evaluating f(x+h) by forgetting to distribute the 2 inside the square root. Write sqrt(2x+2h+5), not sqrt(2x+h+5).

For cube roots, the conjugate trick changes. You need a^3 - b^3 = (a-b)(a^2 + ab + b^2). Let a = cube_root(x+h), b = cube_root(x). Multiply top and bottom by a^2 + ab + b^2.

For f(x) = cube_root(x), the difference quotient is (cube_root(x+h) - cube_root(x))/h. Multiply top and bottom by (cube_root(x+h)^2 + cube_root(x+h)*cube_root(x) + cube_root(x)^2). Top becomes (x+h)-x = h. Cancel h. The simplified form is 1/(cube_root(x+h)^2 + cube_root(x+h)*cube_root(x) + cube_root(x)^2). Limit as h->0 gives 1/(3*cube_root(x)^2).

Most sources skip cube roots entirely. They require the sum/difference of cubes factorization, not the conjugate. The pair to remember: square root uses conjugate (difference of squares); cube root uses the trinomial factor (difference of cubes).

• Not rationalizing the top. You get (sqrt(x+h)-sqrt(x))/h and stop. The h does not cancel without the conjugate.
• Distributing 1/h before combining. Writing (sqrt(x+h)-sqrt(x))/h as sqrt(x+h)/h - sqrt(x)/h. This makes rationalizing impossible and is algebraically incorrect.
• Incorrect conjugate. Using sqrt(x+h)-sqrt(x) as the conjugate of itself. The correct multiplier is sqrt(x+h)+sqrt(x).
• Forgetting to multiply the bottom by the conjugate. You get a top of h but a bottom of h, so the cancellation still works, but the expression is not valid algebraically if you only multiply the top.
• Cancelling before the h factors out. Trying to cancel a single sqrt(x+h) against a term in the bottom. The h is a factor, not a term.

The most common failure is the first one: not attempting the conjugate at all. Students treat the difference quotient sqrt x as something that cannot be simplified and move on. It can be simplified, and the conjugate is how.

Radical Type	Factor Needed	Resulting Numerator	Example f(x)
Square root	Conjugate (a+b)	(x+h)-x = h	sqrt(x)
Cube root	a^2 + ab + b^2	(x+h)-x = h	cube_root(x)
Fourth root	a^3 + a^2b + ab^2 + b^3	(x+h)-x = h	fourth_root(x)

Simplify the difference quotient for each function. Then take the limit as h->0 to find the derivative.

1. f(x) = sqrt(x+1)
2. f(x) = sqrt(3x)
3. f(x) = sqrt(x^2 + 4)
4. f(x) = sqrt(x) + 2
5. f(x) = sqrt(5-2x)

Answers:

1. 1/(2sqrt(x+1))
2. 3/(2sqrt(3x))
3. x/(sqrt(x^2+4))
4. 1/(2sqrt(x))
5. -1/(sqrt(5-2x))

If you got the sign wrong on the last one, check that f(x+h)-f(x) for a decreasing function produces a negative top before the cancellation.