Why Check the Endpoint Separately?
A power-rule formula can correctly describe a derivative at every interior point of a function’s domain and still fail to describe what happens at a boundary. At an endpoint, the function may only be defined on one side, so the derivative must be tested with a one-sided difference quotient. That quotient may have a finite limit, diverge, or not be available at all because the function is undefined at the boundary.
This distinction is especially important for radical and reciprocal-power functions. For example, \(\sqrt{x}\) is defined for \(x\ge 0\), but its familiar interior derivative formula contains \(1/\sqrt{x}\), which is undefined at \(x=0\). The formula alone does not decide whether a finite one-sided derivative exists there. The difference quotient does.
One-Sided Derivatives at Included Endpoints
For a function whose domain contains an interval \([a,b]\), the left endpoint \(a\) has nearby domain values to its right, while the right endpoint \(b\) has nearby domain values to its left. A one-sided derivative measures the limiting slope using only those points in the domain.
The sign restriction on \(h\) is essential: it keeps \(a+h\), or \(b+h\), inside the domain. A limit of \(+\infty\) or \(-\infty\) is not a finite derivative. Nor can a derivative be assigned at a point where the function itself is not defined.
Why the Interior Power Rule Is Not Enough
For \(x>0\), the power rule gives \(\frac{d}{dx}x^p=px^{p-1}\) for the relevant real exponents \(p\). The condition \(x>0\) matters when the expression involves a fractional power. At \(x=0\), the interior formula may be undefined, and even when its values approach a finite number as \(x\to0^+\), that limit alone does not prove the endpoint derivative. The endpoint derivative is defined by a difference quotient based on \(f(0)\).
For a positive fractional power, that quotient has a particularly clear form. For \(f(x)=x^p\), with \(p>0\), \(f(0)=0\), and \(h>0\), so
This expression reveals the endpoint behavior directly. If \(p>1\), then \(h^{p-1}\to0\), so the right-hand derivative is \(0\). If \(p=1\), the quotient equals \(1\). If \(0<p<1\), then \(h^{p-1}=1/h^{1-p}\to+\infty\), so there is no finite right-hand derivative. The result is about the included endpoint \(0\); it does not change the power rule at positive interior points.
Worked Example: The Square-Root Endpoint
Worked Example: The Square-Root Endpoint
Determine whether \(f(x)=\sqrt{x}\), with domain \(x\ge0\), has a finite right-hand derivative at \(x=0\).
First, \(f(0)=\sqrt0=0\). The endpoint quotient must use \(h>0\), because the domain does not include negative \(x\)-values. Thus
As \(h\to0^+\), \(\sqrt h\to0^+\), so \(1/\sqrt h\to+\infty\). Therefore the quotient has no finite limit, and \(f\) has no finite right-hand derivative at \(0\).
Check: The interior power rule gives \(f'(x)=\frac12x^{-1/2}=\frac{1}{2\sqrt{x}}\) for \(x>0\). This formula is undefined at \(0\), consistent with the difference-quotient result. The quotient, not merely the formula’s appearance, establishes the endpoint conclusion.
Worked Example: A Fractional Power with a Finite Endpoint Derivative
Worked Example: A Fractional Power with a Finite Endpoint Derivative
Find the right-hand derivative at \(0\) for \(f(x)=x^{3/2}\), defined for \(x\ge0\).
Since \(f(0)=0\), use \(h>0\) in the endpoint quotient:
As \(h\to0^+\), \(\sqrt h\to0\). Therefore \(f'_+(0)=0\).
Check: At an interior point \(x>0\), the power rule gives \(f'(x)=\frac32x^{1/2}=\frac32\sqrt{x}\). This expression approaches \(0\) as \(x\to0^+\), agreeing with the endpoint quotient. However, the endpoint derivative was established directly from its definition, not by substituting \(x=0\) into an interior formula.
Radicals at a Right Endpoint
The same test applies at a right endpoint, but \(h\) must be negative. This sign change can affect whether a quotient tends to positive or negative infinity. The function’s domain determines which one-sided limit is relevant.
Worked Example: A Radical at the Right Endpoint
Let \(f(x)=\sqrt{4-x}\) on \(0\le x\le4\). Test for a finite left-hand derivative at \(x=4\).
At the endpoint, \(f(4)=\sqrt{4-4}=0\). Since nearby domain values lie to the left, take \(h<0\). Then \(4+h\le4\), and
Because \(h<0\), write \(h=-k\) with \(k>0\). The quotient becomes \(\frac{\sqrt{k}}{-k}=-\frac{1}{\sqrt{k}}\), which tends to \(-\infty\) as \(k\to0^+\). Thus there is no finite left-hand derivative at \(4\).
Check: For \(x<4\), the interior power rule gives \(f'(x)=-\frac{1}{2\sqrt{4-x}}\). As \(x\to4^-\), this tends to \(-\infty\), matching the sign and unbounded behavior found from the endpoint quotient.
Reciprocal Powers and Excluded Boundaries
A reciprocal power can have a different issue: the boundary may not belong to the domain at all. For example, \(x^{-1/2}=1/\sqrt{x}\) is defined for \(x>0\), but it is undefined at \(x=0\). There is no value \(f(0)\) to use in a difference quotient, so there is no derivative at \(0\). The fact that \(0\) is a boundary of the domain does not make it an included endpoint of the function.
On the other hand, a reciprocal expression involving a radical can be defined at the radical’s boundary if its denominator stays nonzero there. In that case, the endpoint quotient must still be checked.
Worked Example: A Reciprocal Radical Defined at the Boundary
Let \(g(x)=\frac{1}{1+\sqrt{x}}\) for \(x\ge0\). Determine whether \(g\) has a finite right-hand derivative at \(0\).
First, \(g(0)=\frac{1}{1+0}=1\). For \(h>0\),
As \(h\to0^+\), the denominator \(\sqrt h(1+\sqrt h)\to0^+\), so the quotient tends to \(-\infty\). Therefore \(g\) has no finite right-hand derivative at \(0\).
Check: For \(x>0\), the quotient rule and the interior derivative of \(\sqrt{x}\) give \(g'(x)=-\frac{1}{2\sqrt{x}(1+\sqrt{x})^2}\). The formula is defined for \(x>0\) but not at \(0\), and its values tend to \(-\infty\) there, consistent with the endpoint quotient. This does not replace the endpoint test.
A Reliable Endpoint Test
When a radical or reciprocal-power expression is being differentiated at a boundary, use the definition before attempting to extend an interior formula. The following sequence keeps the domain and the limit direction explicit.
Decide whether the boundary point is included and whether the function has a value there.
Use \(h\to0^+\) at a left endpoint or \(h\to0^-\) at a right endpoint.
Substitute the endpoint value into \(\frac{f(a+h)-f(a)}{h}\), or the corresponding right-endpoint quotient.
A finite real limit is a one-sided derivative; an infinite limit is not. If the endpoint is excluded, no derivative there is defined.
Common Mistake: Substituting into the Interior Formula
A tempting but invalid shortcut is to take the interior formula for \(\sqrt{x}\), namely \(f'(x)=1/(2\sqrt{x})\), and substitute \(x=0\). That expression is undefined at \(0\), but this alone is not a complete endpoint test: an interior formula’s domain does not determine the endpoint limit of the difference quotient.
For instance, \(x^{3/2}\) also has an interior power-rule formula that is stated for \(x>0\). The formula is not automatically an endpoint formula, yet its endpoint derivative exists: the quotient is \(\sqrt h\), whose limit is \(0\). The correction is to use the endpoint’s one-sided difference quotient, not to assume that the interior rule either must work or must fail at the boundary.
Check Your Understanding
For each question, state the relevant domain and use the appropriate endpoint or domain reasoning.
- Does \(f(x)=x^{2/3}\), on \(x\ge0\), have a finite right-hand derivative at \(0\)? Write its difference quotient and evaluate the limit.
- For \(f(x)=\sqrt{x-2}\), identify the included endpoint and determine whether a finite right-hand derivative exists there.
- Why is \(x^{-1/2}\) not differentiable at \(0\) as a function on its real domain?
- For \(g(x)=1/(2+\sqrt{x})\), write the right-hand difference quotient at \(0\). Does it have a finite limit?
- At a right endpoint, which direction must \(h\) approach from, and why?
