WordGPT

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

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

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

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) / cos(x) and its derivative-3-2-1123-300-200-100100200slope 10.83f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x) / cos(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 10.829.

How to work it out

  1. 1

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

    f(x)=tan⁡(x)cos⁡(x)f(x) = \frac{\tan\left(x\right)}{\cos\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=cos⁡(x)v = \cos\left(x\right).

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

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

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

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

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

xf(x)f'(x)
-2-5.2506463-25.348741
-1-2.882474710.829192
0.50.622508371.8196497
12.882474710.829192
25.2506463-25.348741
30.1439875-1.0511585

Questions

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

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

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

(((-(sin(x) * cos(x)^2) - cos(x) * 2cos(x) * -sin(x)) / (cos(x)^2)^2 - (-sin(x) / cos(x)^2 + tan(x) * -cos(x))) * cos(x)^2 - (cos(x) / cos(x)^2 - tan(x) * -sin(x)) * 2cos(x) * -sin(x)) / (cos(x)^2)^2.

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

The quotient rule.

What is the slope of tan(x) / cos(x) at x = 1?

10.829192.

Related derivatives

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