Why the Inner Derivative Matters
A square root or fractional power can make a function look different from the polynomial expressions you have differentiated so far. The underlying method is familiar: identify the outer function and the inner function, then use the chain rule. The important detail is to multiply by the derivative of the entire inner function.
For example, \(\sqrt{1+x^2}\) is the square-root function applied to \(1+x^2\). Differentiating only the square root’s outer layer would miss how quickly \(1+x^2\) changes with \(x\).
The Rule for Radicals and Fractional Powers
For a positive input \(u\), a fractional power such as \(u^{1/2}\) is defined using the positive square root: \(u^{1/2}=\sqrt{u}\). More generally, the usual real fractional powers are defined on positive inputs, with their meaning determined by the corresponding roots and powers. In particular, \(u^{3/2}=(\sqrt{u})^3\) when \(u\geq 0\).
The power rule says that the derivative of \(u^r\), with respect to \(u\), is \(r u^{r-1}\) when \(u>0\). If the input \(u\) is itself a differentiable function \(g(x)\), the chain rule supplies the additional factor \(g'(x)\).
For a square root, take \(r=\frac12\). Since \(\frac12-1=-\frac12\), the rule gives the equivalent forms below, provided \(g(x)>0\):
The denominator makes clear why this formula requires \(g(x)>0\): it would be zero if \(g(x)=0\). A function involving a square root may still have a derivative at some point where its radicand is zero, but that requires a separate check. The displayed formula alone does not settle what happens there.
Why the Formula Works
Set \(H(u)=u^r\), so that the composite function is \(H(g(x))\). At a point where \(g(x)>0\), the outer function is differentiable and has derivative \(H'(u)=r u^{r-1}\). The chain rule states that the rate of change of a composite function is the outer derivative evaluated at the inner value, multiplied by the inner derivative.
This multiplication reflects how small changes pass through the two layers. A small change in \(x\) changes \(g(x)\) at a rate of \(g'(x)\); the outer function then changes at a rate of \(H'(g(x))\) per unit change in its input. Multiplying the rates gives the rate with respect to \(x\).
Write the expression inside the radical or fractional power as \(g(x)\).
Multiply by its exponent and reduce the exponent by one.
Differentiate all of \(g(x)\) and include \(g'(x)\) as a factor.
Make sure the original function is defined and that the formula’s conditions hold at the point in question.
Worked Examples
Worked Example: Differentiate a Square Root of a Quadratic
Find the derivative of \(y=\sqrt{1+x^2}\), and evaluate it at \(x=2\).
Let \(g(x)=1+x^2\). Then \(g'(x)=2x\). The radicand is positive for every real \(x\), since \(1+x^2\geq 1\), so the square-root derivative formula applies everywhere.
At \(x=2\), the radicand is \(1+2^2=5\), so \(y'(2)=\frac{2}{\sqrt{5}}\).
Check: The outer derivative contributes \(\frac{1}{2\sqrt{1+x^2}}\), and the inner derivative contributes \(2x\). Their product is \(\frac{2x}{2\sqrt{1+x^2}}=\frac{x}{\sqrt{1+x^2}}\), as found. At \(x=2\), these factors give \(\frac{1}{2\sqrt{5}}\cdot 4=\frac{2}{\sqrt{5}}\).
Worked Example: Differentiate a Three-Halves Power
Find the derivative of \(y=(2x-1)^{3/2}\), and state where the derivative formula applies.
For a real-valued function, the base must satisfy \(2x-1\geq 0\), so the function’s domain is \(x\geq\frac12\). For \(x>\frac12\), the base is positive and the fractional power rule applies. Let \(g(x)=2x-1\), so \(g'(x)=2\).
At \(x=2\), the inner value is \(2(2)-1=3\), so \(y'(2)=3\sqrt{3}\).
Check: The outer derivative factor is \(\frac32(2x-1)^{1/2}\), and the inner derivative factor is \(2\). Their product simplifies to \(3\sqrt{2x-1}\); at \(x=2\), it is \(3\sqrt3\).
At the endpoint \(x=\frac12\), the formula’s derivation using a positive base does not apply. However, the right-hand difference quotient is $$ \frac{(2(\frac12+h)-1)^{3/2}-(2(\frac12)-1)^{3/2}}{h} = \frac{(2h)^{3/2}}{h} = 2^{3/2}\sqrt{h}, \qquad h>0. $$ As \(h\) approaches \(0\) from the right, this quotient approaches \(0\). Thus the right-hand derivative at the endpoint is \(0\), even though the displayed formula was initially established only for \(x>\frac12\).
Worked Example: Differentiate a Square Root with a Nonlinear Inner Function
Find the derivative of \(y=\sqrt{3x^2+4x+5}\).
Let \(g(x)=3x^2+4x+5\). Its derivative is \(g'(x)=6x+4\). The radicand is positive for every real \(x\), because $$ 3x^2+4x+5 = 3\left(x+\frac23\right)^2+\frac{11}{3}>0. $$ Therefore, the square-root formula applies everywhere.
For a numerical check at \(x=1\), the radicand is \(3+4+5=12\), and the inner derivative is \(6+4=10\). The chain-rule calculation gives $$ \frac{10}{2\sqrt{12}} = \frac{5}{\sqrt{12}} = \frac{5}{2\sqrt{3}}. $$ The simplified derivative gives the same value: \(\frac{3(1)+2}{\sqrt{12}}=\frac{5}{\sqrt{12}}\).
Check: Before simplifying, the outer factor is \(\frac{1}{2\sqrt{3x^2+4x+5}}\) and the inner factor is \(6x+4\). Their product is the derivative shown.
Worked Example: Differentiate a Five-Halves Power
Find the derivative of \(y=(x^2+4)^{5/2}\), and evaluate it at \(x=1\).
The base \(x^2+4\) is always positive. Set \(g(x)=x^2+4\), so \(g'(x)=2x\). Apply the fractional power rule:
At \(x=1\), \(x^2+4=5\), so \(y'(1)=5\cdot1\cdot5^{3/2}=25\sqrt5\).
Check: Substituting \(x=1\) into the unsimplified chain-rule product gives \(\frac52\cdot5^{3/2}\cdot2=5\cdot5^{3/2}=25\sqrt5\), the same result.
Domain and Endpoint Checks
For an even root such as a square root, the radicand must be nonnegative for the function to be real-valued. The derivative rule using a denominator \(2\sqrt{g(x)}\) requires the radicand to be strictly positive. These are different conditions: the function may be defined at a point where the rule’s denominator is zero.
For example, \(\sqrt{x^2-4}\) is defined when \(x^2-4\geq0\), or \(x\leq-2\) or \(x\geq2\). On the intervals where \(x^2-4>0\), its derivative is $$ \frac{2x}{2\sqrt{x^2-4}} = \frac{x}{\sqrt{x^2-4}}. $$ At \(x=2\) and \(x=-2\), this formula has a zero denominator, so it does not give a finite derivative there. Checking the domain and the point separately prevents applying a valid interior formula where its conditions fail.
A Common Mistake: Omitting the Inner Derivative
A frequent error is to differentiate the square-root layer but stop before multiplying by the derivative of the radicand. For \(y=\sqrt{1+x^2}\), an incorrect calculation is $$ y'=\frac{1}{2\sqrt{1+x^2}}. $$ This treats the inner expression \(1+x^2\) as though its derivative were \(1\). In fact, its derivative is \(2x\).
At \(x=1\), the incorrect result is \(\frac{1}{2\sqrt2}\), while the correct result is \(\frac{1}{\sqrt2}\). The missing factor of \(2x\) changes the answer. A reliable check is to name the inner function explicitly and write its derivative before simplifying.
Why This Skill Matters
Radicals and fractional powers often appear inside longer expressions. Differentiating them correctly depends on seeing the layers clearly, not on whether the inner function is a simple linear expression or a more complicated polynomial. Keeping the inner derivative visible until the final simplification makes errors easier to spot and gives a correct starting point for more involved composite derivatives.
Check Your Understanding
For each question, identify the inner function and check the domain before differentiating.
- Differentiate \(y=\sqrt{5x+4}\). For which \(x\)-values does the derivative formula apply?
- Differentiate \(y=(x^2+1)^{3/2}\), and evaluate the derivative at \(x=1\).
- Differentiate \(y=\sqrt{4-x^2}\). Where is the function defined, and where does the square-root derivative formula apply?
- A student differentiates \((3x+2)^{5/2}\) and writes \(\frac52(3x+2)^{3/2}\). What factor is missing, and what is the corrected derivative?
- Explain why a function involving a square root can be defined at a point where the standard square-root derivative formula does not apply.
