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

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

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

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Graph of f(x) = tan(x) and its derivative-3-2-1123-20204060slope 3.426f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 3.4255.

How to work it out

  1. 1

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

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

    This is a standard derivative from the table.

  3. 3

    Simplify:

    f′(x)=1cos⁡(x)2f'(x) = \frac{1}{\cos\left(x\right)^{2}}
  4. 4

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

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

Values of f and f' for tan(x)

xf(x)f'(x)
-22.18503995.7743992
-1-1.55740773.4255188
0.50.546302491.2984464
11.55740773.4255188
2-2.18503995.7743992
3-0.142546541.0203195

Questions

What is the derivative of tan(x)?

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

What is the second derivative of tan(x)?

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

Which rule do you use for tan(x)?

The standard derivative.

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

3.4255188.

Related derivatives

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