Why Conjugates Reveal the Slope
A square-root function can look difficult to differentiate from first principles because its difference quotient contains a radical in the numerator. Multiplying by a conjugate changes that radical difference into an ordinary algebraic difference. The resulting expression is simpler, and its limit gives the slope.
The input restriction matters. The real-valued function \(f(x)=\sqrt{x}\) is defined for \(x\ge 0\), but an ordinary two-sided derivative at an input requires the function to be defined on an open interval around that input. Thus this tutorial finds the derivative at positive inputs, which are interior points of the domain.
Here \(h\) is the change in input, and the quotient measures the change in output per unit of input change. To use the definition, both \(a\) and \(a+h\) must be in the function's domain. At an interior point, all sufficiently small positive and negative values of \(h\) are allowed.
The Square-Root Function at a Positive Input
Let \(f(x)=\sqrt{x}\), and fix \(a>0\). Since \(a\) is positive, there is an open interval around \(a\) on which the square root is defined. For small enough nonzero \(h\), the difference quotient is
Direct substitution of \(h=0\) would produce \(0/0\), which is not a value of the quotient. Instead, multiply the numerator and denominator by the conjugate \(\sqrt{a+h}+\sqrt{a}\). This is permitted for sufficiently small \(h\), because \(a+h>0\) and the conjugate is then positive.
The cancellation of \(h\) is valid because the difference quotient is defined only for \(h\ne 0\). After cancellation, the expression has a limit as \(h\to 0\). The square-root function is continuous at positive inputs, so \(\sqrt{a+h}\to\sqrt{a}\). The denominator therefore tends to \(2\sqrt{a}\), which is nonzero because \(a>0\).
Worked Examples
Worked Example: Finding the Slope of \(\sqrt{x}\) at \(x=9\)
Use the definition at \(a=9\). The function value is \(f(9)=\sqrt{9}=3\). For \(h\ne 0\) sufficiently close to zero,
For example, if \(|h|<1\), then \(9+h>8>0\), so the quotient is defined for both signs of \(h\). Taking the limit gives
Check the limiting expression: as \(h\to0\), its denominator approaches \(6\), so its value approaches \(1/6\), as stated. The slope of \(y=\sqrt{x}\) at \(x=9\) is therefore \(1/6\).
The same algebra works when the quantity under the square root is an affine function of \(x\). The change in that quantity must be retained: replacing \(x\) by \(x+h\) changes \(mx+b\) by \(mh\), not merely by \(h\).
To justify the formula directly, write \(A=ma+b\), so \(A>0\). For nonzero \(h\) sufficiently close to zero, \(A+mh>0\), and the definition gives
As \(h\to0\), the denominator tends to \(2\sqrt{A}\), which is positive. Hence the quotient tends to \(m/(2\sqrt{A})\), or \(m/(2\sqrt{ma+b})\). This derivation also covers \(m=0\) when \(b>0\): the quotient is then zero, in agreement with the formula.
Worked Example: Differentiating \(\sqrt{2x+3}\) at \(x=3\)
Let \(f(x)=\sqrt{2x+3}\). Its real domain satisfies \(2x+3\ge0\), or \(x\ge-3/2\). Since \(3>-3/2\), the input \(3\) is interior. Also, \(f(3)=\sqrt{9}=3\). For \(h\ne0\) close enough to zero,
For \(|h|<1\), \(9+2h>7>0\), so this expression is defined from both sides of zero. Therefore
Checking the simplified quotient, its numerator is \(2\) and its denominator tends to \(6\); its limit is \(2/6=1/3\). The factor \(2\) comes from the change \(2(3+h)-2(3)=2h\).
Worked Example: A Decreasing Square-Root Function
Let \(g(x)=\sqrt{5-x}\), and find its derivative at \(x=1\) from the definition. The real domain is \(x\le5\), so \(1\) is interior. Since \(g(1)=\sqrt{4}=2\),
For \(|h|<1\), \(4-h>3>0\), so both positive and negative \(h\) near zero are allowed. Thus
The negative sign is consistent with the quotient: increasing \(x\) decreases \(5-x\), and therefore decreases the square-root output. The simplified expression tends to \(-1/4\), confirming the computed derivative.
Restrictions and a Common Error
There are two restrictions to keep visible during these calculations. First, the radicands must remain nonnegative for the function values to be real. Second, the derivative definition uses a two-sided limit at an interior point, so all sufficiently small positive and negative increments must be allowed. For \(f(x)=\sqrt{x}\), this is true at every \(a>0\), but not at \(a=0\). The formula derived here is therefore not a claim about a derivative at \(0\); that point is an endpoint of the real domain.
For an affine radicand \(mx+b\), the condition \(ma+b>0\) ensures an interval around \(a\) on which the radicand stays positive. If \(m\ne0\), the point where \(ma+b=0\) is a boundary of the domain, not an interior point. The strict inequality also guarantees that the denominator in the derivative formula is nonzero.
Why the Method Works
The conjugate step uses the identity \((u-v)(u+v)=u^2-v^2\). With \(u=\sqrt{a+h}\) and \(v=\sqrt{a}\), it converts the numerator into \((a+h)-a=h\). That factor cancels the \(h\) in the difference quotient, leaving an expression whose denominator has a finite, nonzero limit. This is why rationalizing is more than a cosmetic rearrangement: it exposes the limit that the original form concealed.
For \(\sqrt{mx+b}\), the same identity converts the numerator into the change in the radicand, \(mh\). Dividing by the input change \(h\) leaves \(m\), which accounts for the coefficient in the derivative. No differentiation rule for compositions is needed; the result follows directly from the definition and the conjugate identity.
Summary
At a positive input \(a\), the difference quotient for \(\sqrt{x}\) can be rationalized and simplified to \(1/(\sqrt{a+h}+\sqrt{a})\). Its limit is \(1/(2\sqrt{a})\). More generally, if the radicand is \(mx+b\) and is strictly positive at the input, the derivative is \(m/(2\sqrt{mx+b})\) there. The strict positivity condition identifies an interior point and keeps the limiting denominator nonzero.
Check Your Understanding
Use the derivative definition and the conjugate method where appropriate.
- Why must the input \(a\) be positive for the ordinary two-sided derivative of \(f(x)=\sqrt{x}\) considered here?
- Find \(f'(4)\) for \(f(x)=\sqrt{x}\) by rationalizing the difference quotient.
- For \(p(x)=\sqrt{3x+1}\), identify its real domain and determine whether \(x=2\) is an interior point.
- Use the definition to find the derivative of \(p(x)=\sqrt{3x+1}\) at \(x=2\). Show where the factor \(3\) appears.
- For \(q(x)=\sqrt{7-x}\), explain why the derivative at \(x=2\) is negative and calculate it from the affine-radicand result.
