WordGPT

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

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

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

Type any function of x; the derivative is computed symbolically in your browser.

Explain it step by step in chatMake a short video of it
Graph of f(x) = tan(x) * ln(x) and its derivative-3-2-1123-55101520slope 1.557f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x) * ln(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 1.5574.

How to work it out

  1. 1

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

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

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \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)
  5. 5

    Simplify:

    f′(x)=ln⁡(x)cos⁡(x)2+tan⁡(x)1xf'(x) = \frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{x}
  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)−1x2f''(x) = \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}} + \frac{\frac{1}{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{-1}{x^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\tan\left(x\right) \ln\left(x\right)\right] = \frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{x} \]
\[ f(x) = \tan\left(x\right) \ln\left(x\right) \]
\[ u = \tan\left(x\right),\quad v = \ln\left(x\right) \]
\[ u' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = \frac{1}{x} \]
\[ f'(x) = \left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(\ln\left(x\right)\right) + \left(\tan\left(x\right)\right)\left(\frac{1}{x}\right) \]
\[ f'(x) = \frac{\ln\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{x} \]
\[ f''(x) = \frac{\frac{\cos\left(x\right)^{2}}{x} - \ln\left(x\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} + \frac{\frac{1}{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{-1}{x^{2}} \]

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

xf(x)f'(x)
0.5-0.378668030.19259051
101.5574077
2-1.51455422.9099886
3-0.156603381.07342

Questions

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

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

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}.

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

The product rule.

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

1.5574077.

Related derivatives

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