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Powers, roots and polynomials

Derivative of 1/x: Formula, Proof and Examples

The derivative of 1/x is -1/x². See the limit definition proof, the power rule with x⁻¹, the chain rule for 1/u and worked examples like 1/(2x + 5).

Powers, roots and polynomials Updated 2 min read

At a glance

\frac{d}{dx}\,\frac{1}{x} = -\frac{1}{x^2}
Derivative
-\dfrac{1}{x^2} for x \ne 0
As a power
\dfrac{d}{dx}\,x^{-1} = -x^{-2}
Chain rule form
\dfrac{d}{dx}\,\dfrac{1}{u} = -\dfrac{u'}{u^2}
Sign
Always negative, so \frac{1}{x} decreases on each side of 0
Integral version
\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C
On this page
  1. Graph of 1/x and its derivative
  2. Proof from the limit definition
  3. Proof with the power rule
  4. The chain rule version: derivative of 1/u
  5. Worked examples
  6. Common mistakes

The derivative of \dfrac{1}{x} is -\dfrac{1}{x^2}. The minus sign tells you the curve is always going downhill, on both sides of the vertical asymptote.

\frac{d}{dx}\,\frac{1}{x} = -\frac{1}{x^2}, \qquad x \ne 0

Graph of 1/x and its derivative

Graph of y = 1/x and its derivative y = -1/x²
Both branches of \frac{1}{x} fall from left to right, so the derivative is negative everywhere it exists.

Near x = 0 the branches are nearly vertical, and -\frac{1}{x^2} is a huge negative number. Far from 0 the curve flattens, and -\frac{1}{x^2} approaches 0.

Proof from the limit definition

f'(x) = \lim_{h \to 0}\frac{\frac{1}{x + h} - \frac{1}{x}}{h}

Combine the fractions on top over the common denominator x(x + h):

\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}

Divide by h and let h \to 0:

f'(x) = \lim_{h \to 0}\frac{-1}{x(x + h)} = -\frac{1}{x^2}

Proof with the power rule

Since \frac{1}{x} = x^{-1}, the power rule gives -1 \cdot x^{-2} = -\frac{1}{x^2}. The quotient rule agrees: \frac{0 \cdot x - 1 \cdot 1}{x^2} = -\frac{1}{x^2}.

The chain rule version: derivative of 1/u

\frac{d}{dx}\,\frac{1}{u} = -\frac{1}{u^2}\cdot u' = -\frac{u'}{u^2}
Function Inside u Derivative
\frac{1}{2x + 5} 2x + 5 -\frac{2}{(2x + 5)^2}
\frac{1}{x^2 + 1} x^2 + 1 -\frac{2x}{(x^2 + 1)^2}
\frac{1}{\sin x} \sin x -\frac{\cos x}{\sin^2 x}
\frac{1}{e^x + 1} e^x + 1 -\frac{e^x}{(e^x + 1)^2}

The third row is the same as -\csc x\cot x, the derivative of csc x.

Worked examples

Example 1: a sum. \frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2}. Setting this to 0 gives x = \pm 1, where x + \frac{1}{x} has its turning points.

Example 2: a tangent line. At x = 2, y = \frac{1}{2} and the slope is -\frac{1}{4}:

y - \frac{1}{2} = -\frac{1}{4}(x - 2) \quad\Longrightarrow\quad y = -\frac{x}{4} + 1

Example 3: second derivative. \frac{d}{dx}\left(-x^{-2}\right) = 2x^{-3} = \frac{2}{x^3}. It is positive for x > 0 and negative for x < 0, so the right branch is concave up and the left branch concave down.

Common mistakes

  1. Answering \ln x or \frac{1}{x^2}. The derivative is -\frac{1}{x^2}. The function \ln x is the antiderivative, not the derivative.
  2. Losing the minus sign. x^{-1} brings down -1.
  3. Writing x^0. Lowering -1 by one gives -2, not 0.
  4. Forgetting u'. \frac{d}{dx}\frac{1}{3x} is -\frac{3}{(3x)^2} = -\frac{1}{3x^2}, not -\frac{1}{9x^2}.

The integral side is on the integral of 1/x page, and the curve itself is on the reciprocal function page.

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Frequently asked questions

What is the derivative of 1/x?

The derivative of \frac{1}{x} is -\frac{1}{x^2}, for every x \ne 0.

What is the derivative of 1/(2x)?

Write it as \frac{1}{2}x^{-1}. The derivative is -\frac{1}{2}x^{-2} = -\frac{1}{2x^2}.

What is the derivative of 5/x?

It is 5 \cdot \left(-\frac{1}{x^2}\right) = -\frac{5}{x^2}.

Which function has derivative 1/x?

The natural log: \frac{d}{dx}\ln x = \frac{1}{x}. That is why \int \frac{1}{x}\,dx = \ln\lvert x\rvert + C.

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