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Basic Differentiation Rules · Tutorial 449 of 1495

Mixed Powers With Roots And Reciprocals

Rewrite roots and reciprocals as powers, then differentiate each term while keeping the function’s domain in view.

Beginner 10 min read

What You'll Learn

  • Rewrite square roots and reciprocal powers as fractional and negative exponents
  • Find the domain of a mixed expression before differentiating
  • Apply the power rule term by term when its conditions hold
  • Keep coefficients and signs correct when lowering exponents
  • Distinguish ordinary derivatives from behavior at domain endpoints
  • Check a mixed derivative using the definition at a convenient point

Why Rewrite Roots and Reciprocals?

A formula containing a polynomial term, a root, and a fraction can look like several different kinds of calculation. For example, a square root is written with a radical sign, while a reciprocal is written with a denominator. Rewriting both as powers puts the terms into a common form, so the power rule can be applied consistently.

The rewriting is useful only when the domain is tracked correctly. A root may restrict which inputs are allowed, and a reciprocal excludes inputs that make its denominator zero. The derivative formula must also be used only where its powers and expressions are defined.

Key idea: First identify where the original function is defined. Then rewrite each root or reciprocal as a power, differentiate each term where the power rule applies, and express the resulting derivative on its valid domain.

Rewriting Roots and Reciprocals

For \(x>0\), roots and reciprocal powers can be written using fractional and negative exponents. In particular, \(\sqrt{x}=x^{1/2}\), and a reciprocal such as \(1/x^n\) is \(x^{-n}\), provided \(x\ne 0\). The same rule applies to a reciprocal of a root: \(1/\sqrt{x}=x^{-1/2}\) for \(x>0\).

$$ \sqrt{x}=x^{1/2},\qquad \frac{1}{x^n}=x^{-n},\qquad \frac{1}{\sqrt{x}}=x^{-1/2}\quad (x>0). $$

These equalities make the exponent rules available, but they do not remove the original domain restrictions. For example, \(1/\sqrt{x}\) requires a real square root and a nonzero denominator. Together these conditions require \(x>0\), not merely \(x\ge 0\).

Definition: The domain of a real-valued function is the set of real inputs for which its defining expression has a real value. When differentiating a function, first determine this domain; a derivative formula can only be used at points where the function is defined and differentiable.

The Power Rule and Its Conditions

The power rule says that for every real exponent \(r\), if \(x>0\), then \(x^r\) is differentiable and its derivative is \(r x^{r-1}\). At points outside \(x>0\), use the formula only where the real powers involved are defined and the expression \(r x^{r-1}\) is defined. In particular, do not substitute \(x=0\) automatically: the derivative at zero must be considered separately if zero belongs to the function’s domain.

$$ \frac{d}{dx}x^r=r x^{r-1}\qquad (x>0,\; r\in\mathbb R). $$

For instance, \(x^0=1\) is a constant function, so its derivative is \(0\), including at \(x=0\). The displayed power-rule expression at \(r=0,\ x=0\) would contain \(0\cdot 0^{-1}\), which is undefined; the formula must not be used there. In the examples below, reciprocal terms or square-root terms restrict the domain to \(x>0\), so the stated power rule applies throughout each domain.

For a finite sum of differentiable terms, the derivative of the sum is the sum of their derivatives. A constant multiple remains a constant multiple after differentiation. These rules let us handle each term separately once the domain and power-rule conditions have been checked.

A Reliable Workflow

1
Find the domain.
Check the radicands and denominators in the original expression. Intersect the restrictions from all terms.
2
Rewrite the terms.
Express roots as fractional powers and reciprocal powers as negative powers. Preserve each coefficient and sign.
3
Differentiate term by term.
For each \(x^r\), multiply by \(r\) and replace the exponent by \(r-1\). Keep any constant multiplier.
4
State where the result applies.
Give the derivative on the interior of the original domain, and do not claim a derivative at an excluded point or an endpoint without checking it separately.

Worked Examples

Worked Example: A Square Root and a Reciprocal

Differentiate \(f(x)=x^2+2\sqrt{x}-3/x\), and check the derivative at \(x=1\).

Step 1: Find the domain. The square root requires \(x\ge 0\). The denominator \(x\) requires \(x\ne 0\). Both conditions together give \(x>0\).

Step 2: Rewrite with exponents.

$$f(x)=x^2+2x^{1/2}-3x^{-1},\qquad x>0.$$

Step 3: Differentiate each term. The derivative of \(x^2\) is \(2x\). The derivative of \(2x^{1/2}\) is \(2(1/2)x^{1/2-1}=x^{-1/2}\). The derivative of \(-3x^{-1}\) is \(-3(-1)x^{-1-1}=3x^{-2}\). Therefore,

$$f'(x)=2x+x^{-1/2}+3x^{-2},\qquad x>0.$$

Check at \(x=1\). The formula gives \(f'(1)=2+1+3=6\). From the difference quotient, \(f(1)=1+2-3=0\), so

$$\frac{f(1+h)-f(1)}{h}=2+h+\frac{2}{\sqrt{1+h}+1}+\frac{3}{1+h}.$$

Here the square-root term follows by multiplying \((\sqrt{1+h}-1)/h\) by its conjugate. Taking \(h\to 0\) gives \(2+1+3=6\), agreeing with the derivative formula.

Worked Example: Several Fractional Powers

Differentiate \(g(x)=5x^{3/2}-2\sqrt{x}+7/\sqrt{x}\), and evaluate the derivative at \(x=4\).

Step 1: Find the domain. The square root requires \(x\ge 0\), and the denominator \(\sqrt{x}\) must be nonzero. Thus the domain is \(x>0\).

Step 2: Rewrite the roots. Since \(\sqrt{x}=x^{1/2}\) and \(1/\sqrt{x}=x^{-1/2}\),

$$g(x)=5x^{3/2}-2x^{1/2}+7x^{-1/2}.$$

Step 3: Apply the power rule. The first term gives \(5(3/2)x^{3/2-1}=(15/2)x^{1/2}\). The second gives \(-2(1/2)x^{1/2-1}=-x^{-1/2}\). The third gives \(7(-1/2)x^{-1/2-1}=-(7/2)x^{-3/2}\). Hence

$$g'(x)=\frac{15}{2}x^{1/2}-x^{-1/2}-\frac{7}{2}x^{-3/2},\qquad x>0.$$

Step 4: Evaluate at \(x=4\). We have \(4^{1/2}=2\), \(4^{-1/2}=1/2\), and \(4^{-3/2}=1/8\). Therefore,

$$g'(4)=\frac{15}{2}(2)-\frac{1}{2}-\frac{7}{2}\left(\frac18\right)=15-\frac12-\frac{7}{16}=\frac{240-8-7}{16}=\frac{225}{16}.$$

Check. Differentiating the rewritten expression term by term gives the three displayed derivative terms with coefficients \(15/2\), \(-1\), and \(-7/2\); substituting \(x=4\) gives \(225/16\). All powers used in this check are defined because \(4>0\).

Worked Example: Polynomial, Root, Reciprocal Square, and Constant

Differentiate \(p(x)=4x^3-6\sqrt{x}+8/x^2+9\), and find \(p'(1)\).

Step 1: Find the domain. The square root requires \(x\ge 0\), while \(x^2\) in the denominator requires \(x\ne 0\). Thus \(p\) is defined for \(x>0\).

Step 2: Rewrite with exponents.

$$p(x)=4x^3-6x^{1/2}+8x^{-2}+9.$$

Step 3: Differentiate each term. The terms give \(4(3)x^2=12x^2\), \(-6(1/2)x^{-1/2}=-3x^{-1/2}\), and \(8(-2)x^{-3}=-16x^{-3}\). The constant \(9\) has derivative \(0\). Therefore,

$$p'(x)=12x^2-3x^{-1/2}-16x^{-3},\qquad x>0.$$

Step 4: Evaluate and check. At \(x=1\), each power equals \(1\), so \(p'(1)=12-3-16=-7\). To check using the difference quotient, \(p(1)=4-6+8+9=15\). For \(h\ne 0\),

$$\frac{p(1+h)-p(1)}{h}=12+12h+4h^2-\frac{6}{\sqrt{1+h}+1}-\frac{8(2+h)}{(1+h)^2}.$$

The square-root quotient is rationalized, and the reciprocal-square difference is combined over a common denominator. As \(h\to0\), the expression tends to \(12-3-16=-7\), matching the formula.

Why the Rewriting Works

Rewriting does not change the function on the domain where the original and rewritten expressions are both defined. For example, if \(x>0\), then \(1/\sqrt{x}=x^{-1/2}\): by the definition of a negative exponent, \(x^{-1/2}=1/x^{1/2}\), and \(x^{1/2}=\sqrt{x}\). The denominator is nonzero because \(x>0\).

Once the expressions agree throughout the domain, their derivatives there can be computed from the same function values. The power rule supplies the derivative of each power, while the sum and constant-multiple rules combine those derivatives. This is why the method is systematic: every term is converted to a form covered by a known rule, without changing the function on the points being considered.

Common Mistake: Losing a Negative Sign

A negative exponent does not mean the derivative keeps a negative coefficient. The exponent itself is multiplied into the coefficient. For instance, the following calculation is wrong:

$$\frac{d}{dx}\left(5x^{-2}\right)\ne 5x^{-3}.$$

The exponent \(-2\) must multiply the coefficient \(5\), and the new exponent is \(-2-1=-3\). Thus the corrected calculation is

$$\frac{d}{dx}\left(5x^{-2}\right)=5(-2)x^{-3}=-10x^{-3},\qquad x>0.$$

Another domain error is to include \(x=0\) merely because a square root is defined there. In an expression containing \(1/\sqrt{x}\), the denominator is zero at \(x=0\), so the original function is not defined there. A derivative formula cannot restore a point that was excluded from the function’s domain.

Domain Endpoints and the Derivative

For expressions involving both roots and reciprocals, the combined domain often becomes \(x>0\). The derivative formulas in this tutorial are stated on that open interval, where the power rule applies to all the displayed real powers. If a function’s domain includes an endpoint such as \(x=0\), do not infer differentiability there from a formula valid for \(x>0\). Check the definition at the endpoint using the appropriate one-sided difference quotient, if an endpoint derivative is under consideration.

This distinction matters because some expressions are defined at zero but have no finite derivative there, while others may have a one-sided derivative. For example, \(\sqrt{x}\) is defined for \(x\ge0\), but its right-hand difference quotient at zero is \(\sqrt{h}/h=1/\sqrt{h}\) for \(h>0\), which does not approach a finite value. In contrast, \(x^0=1\) is constant and has derivative zero; the general power formula should not be substituted at \(x=0\) when it produces an undefined expression.

Check Your Understanding

For each question, state any domain restriction before differentiating. Use the power rule only where its conditions are satisfied.

  1. Rewrite \(3\sqrt{x}+2/x^3\) using exponents and state its domain.
  2. Differentiate \(q(x)=2x^4-5\sqrt{x}+1/x\), and state the domain of the derivative formula.
  3. Differentiate \(r(x)=4x^{-1/2}-3x^{2/3}\) for \(x>0\).
  4. Explain why \(1/\sqrt{x}\) is not defined at \(x=0\), even though \(\sqrt{x}\) is defined there.
  5. What is the derivative of \(x^0=1\) at \(x=0\), and why should the expression \(0\cdot0^{-1}\) not be used to obtain it?

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