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Derivative of ln⁡(−x)\ln\left(-x\right)

The derivative of ln⁡(−x)\ln\left(-x\right) is −1−x\frac{-1}{-x}.

f(x) =
ddx[ln⁡(−x)]=−1−x\frac{d}{dx}\left[\ln\left(-x\right)\right] = \frac{-1}{-x}

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Graph of f(x) = ln(-x) and its derivative-3-2-1123-3-2-112slope -0.5f(x)f'(x)tangent at x = -2
Graph of f(x) = ln(-x) and its derivative. The dashed line is the tangent at x = -2; its slope is f'(-2) = -0.5.

How to work it out

  1. 1

    We differentiate f(x)=ln⁡(−x)f(x) = \ln\left(-x\right) with respect to x. The rule that applies is the chain rule.

    f(x)=ln⁡(−x)f(x) = \ln\left(-x\right)
  2. 2

    Chain rule: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. The inside is g(x)=−xg(x) = -x.

    g(x)=−x,g′(x)=−1g(x) = -x,\quad g'(x) = -1
  3. 3

    The outer function is ln(u); its derivative evaluated at g(x), times g'(x):

    f′(x)=[1u]u=g(x)⋅g′(x)f'(x) = \left[\frac{1}{u}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    f′(x)=−1−xf'(x) = \frac{-1}{-x}
  5. 5

    Differentiating once more gives the second derivative, which tells you about curvature:

    f′′(x)=−1(−x)2f''(x) = \frac{-1}{\left(-x\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\ln\left(-x\right)\right] = \frac{-1}{-x} \]
\[ f(x) = \ln\left(-x\right) \]
\[ g(x) = -x,\quad g'(x) = -1 \]
\[ f'(x) = \left[\frac{1}{u}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = \frac{-1}{-x} \]
\[ f''(x) = \frac{-1}{\left(-x\right)^{2}} \]

Values of f and f' for ln(-x)

xf(x)f'(x)
-20.69314718-0.5
-10-1

Questions

What is the derivative of ln(-x)?

The derivative of ln⁡(−x)\ln\left(-x\right) is −1−x\frac{-1}{-x}.

What is the second derivative of ln(-x)?

-1 / (-x)^2.

Which rule do you use for ln(-x)?

The chain rule.

What is the slope of ln(-x) at x = -2?

-0.5.

Related derivatives

    Derivative of ln(-x): -1 / -x (with steps) | WordGPT