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Derivative of ln⁡(x)arctan⁡(x)\ln\left(x\right) \arctan\left(x\right)

The derivative of ln⁡(x)arctan⁡(x)\ln\left(x\right) \arctan\left(x\right) is arctan⁡(x)x+ln⁡(x)11+x2\frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}}.

f(x) =
ddx[ln⁡(x)arctan⁡(x)]=arctan⁡(x)x+ln⁡(x)11+x2\frac{d}{dx}\left[\ln\left(x\right) \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}}

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Graph of f(x) = ln(x) * atan(x) and its derivative-3-2-1123-3-2-11slope 0.7854f(x)f'(x)tangent at x = 1
Graph of f(x) = ln(x) * atan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 0.7854.

How to work it out

  1. 1

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

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

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u=ln⁡(x)u = \ln\left(x\right) and v=arctan⁡(x)v = \arctan\left(x\right).

    u=ln⁡(x),v=arctan⁡(x)u = \ln\left(x\right),\quad v = \arctan\left(x\right)
  3. 3

    Differentiate each piece:

    u′=1x,v′=11+x2u' = \frac{1}{x},\quad v' = \frac{1}{1 + x^{2}}
  4. 4

    Put them in the rule:

    f′(x)=(1x)(arctan⁡(x))+(ln⁡(x))(11+x2)f'(x) = \left(\frac{1}{x}\right)\left(\arctan\left(x\right)\right) + \left(\ln\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right)
  5. 5

    Simplify:

    f′(x)=arctan⁡(x)x+ln⁡(x)11+x2f'(x) = \frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}}
  6. 6

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

    f′′(x)=x1+x2−arctan⁡(x)x2+11+x2x+ln⁡(x)−(2x)(1+x2)2f''(x) = \frac{\frac{x}{1 + x^{2}} - \arctan\left(x\right)}{x^{2}} + \frac{\frac{1}{1 + x^{2}}}{x} + \ln\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\ln\left(x\right) \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f(x) = \ln\left(x\right) \arctan\left(x\right) \]
\[ u = \ln\left(x\right),\quad v = \arctan\left(x\right) \]
\[ u' = \frac{1}{x},\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \left(\frac{1}{x}\right)\left(\arctan\left(x\right)\right) + \left(\ln\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right) \]
\[ f'(x) = \frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f''(x) = \frac{\frac{x}{1 + x^{2}} - \arctan\left(x\right)}{x^{2}} + \frac{\frac{1}{1 + x^{2}}}{x} + \ln\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}} \]

Values of f and f' for ln(x) * atan(x)

xf(x)f'(x)
0.5-0.321376030.37277747
100.78539816
20.767417010.6922038
31.3722170.52620982

Questions

What is the derivative of ln(x) * atan(x)?

The derivative of ln⁡(x)arctan⁡(x)\ln\left(x\right) \arctan\left(x\right) is arctan⁡(x)x+ln⁡(x)11+x2\frac{\arctan\left(x\right)}{x} + \ln\left(x\right) \frac{1}{1 + x^{2}}.

What is the second derivative of ln(x) * atan(x)?

(x / (1 + x2x^{2}) - atan(x)) / x2x^{2} + 1 / (1 + x2x^{2}) / x + ln(x) * -(2x) / (1 + x2x^{2})^2.

Which rule do you use for ln(x) * atan(x)?

The product rule.

What is the slope of ln(x) * atan(x) at x = 1?

0.78539816.

Related derivatives

    Derivative of ln(x) * atan(x): atan(x) / x + ln(x) * 1 / (1 + x^2) (with steps) | WordGPT