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Derivative of tan⁡(x)ex\frac{\tan\left(x\right)}{e^{x}}

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

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

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Graph of f(x) = tan(x) / exp(x) and its derivative-3-2-1123-2020406080100120slope 0.6872f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x) / exp(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 0.68724.

How to work it out

  1. 1

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

    f(x)=tan⁡(x)exf(x) = \frac{\tan\left(x\right)}{e^{x}}
  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=exv = e^{x}.

    u=tan⁡(x),v=exu = \tan\left(x\right),\quad v = e^{x}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f'(x) = \frac{\frac{e^{x}}{\cos\left(x\right)^{2}} - \tan\left(x\right) e^{x}}{e^{x}^{2}}
  6. 6

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

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

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

xf(x)f'(x)
-216.14538226.521978
-1-4.233473113.544999
0.50.331349210.45619835
10.572938280.68723967
2-0.295712991.0771929
3-0.00709697450.057895692

Questions

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

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

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

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

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

The quotient rule.

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

0.68723967.

Related derivatives

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