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Derivative of tan⁡(x)ln⁡(x)\frac{\tan\left(x\right)}{\ln\left(x\right)}

The derivative of tan⁡(x)ln⁡(x)\frac{\tan\left(x\right)}{\ln\left(x\right)} is ln⁡(x)cos⁡(x)2−tan⁡(x)1xln⁡(x)2\frac{\frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{x}}{\ln\left(x\right)^{2}}.

f(x) =
ddx[tan⁡(x)ln⁡(x)]=ln⁡(x)cos⁡(x)2−tan⁡(x)1xln⁡(x)2\frac{d}{dx}\left[\frac{\tan\left(x\right)}{\ln\left(x\right)}\right] = \frac{\frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{x}}{\ln\left(x\right)^{2}}

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Graph of f(x) = tan(x) / ln(x) and its derivative-3-2-1123-60-40-2020406080100slope -4.147f(x)f'(x)tangent at x = 0.5
Graph of f(x) = tan(x) / ln(x) and its derivative. The dashed line is the tangent at x = 0.5; its slope is f'(0.5) = -4.1474.

How to work it out

  1. 1

    We differentiate f(x)=tan⁡(x)ln⁡(x)f(x) = \frac{\tan\left(x\right)}{\ln\left(x\right)} with respect to x. The rule that applies is the quotient rule.

    f(x)=tan⁡(x)ln⁡(x)f(x) = \frac{\tan\left(x\right)}{\ln\left(x\right)}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=tan⁡(x)u = \tan\left(x\right) and v=ln⁡(x)v = \ln\left(x\right).

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

    Differentiate each piece:

    u′=1cos⁡(x)2,v′=1xu' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = \frac{1}{x}
  4. 4

    Put them in the rule:

    f′(x)=(1cos⁡(x)2)(ln⁡(x))−(tan⁡(x))(1x)(ln⁡(x))2f'(x) = \frac{\left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(\ln\left(x\right)\right) - \left(\tan\left(x\right)\right)\left(\frac{1}{x}\right)}{\left(\ln\left(x\right)\right)^{2}}
  5. 5

    Simplify:

    f′(x)=ln⁡(x)cos⁡(x)2−tan⁡(x)1xln⁡(x)2f'(x) = \frac{\frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{x}}{\ln\left(x\right)^{2}}
  6. 6

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

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

Values of f and f' for tan(x) / ln(x)

xf(x)f'(x)
0.5-0.78814789-4.1473763
2-3.152346210.604634
3-0.129751460.96810314

Questions

What is the derivative of tan(x) / ln(x)?

The derivative of tan⁡(x)ln⁡(x)\frac{\tan\left(x\right)}{\ln\left(x\right)} is ln⁡(x)cos⁡(x)2−tan⁡(x)1xln⁡(x)2\frac{\frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{x}}{\ln\left(x\right)^{2}}.

What is the second derivative of tan(x) / ln(x)?

(((cos(x)^2 / x - ln(x) * 2cos(x) * -sin(x)) / (cos(x)^2)^2 - (1 / x / cos(x)^2 + tan(x) * -1 / x2x^{2})) * ln(x)^2 - (ln(x) / cos(x)^2 - tan(x) * 1 / x) * 2ln(x) * 1 / x) / (ln(x)^2)^2.

Which rule do you use for tan(x) / ln(x)?

The quotient rule.

What is the slope of tan(x) / ln(x) at x = 0.5?

-4.1473763.

Related derivatives

    Derivative of tan(x) / ln(x): (ln(x) / cos(x)^2 - tan(x) * 1 / x) / ln(x)^2 (with steps) | WordGPT