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

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

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

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Graph of f(x) = tan(x) * exp(x) and its derivative-3-2-1123-60-40-2020406080100120slope 13.54f(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) = 13.545.

How to work it out

  1. 1

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

    f(x)=tan⁡(x)exf(x) = \tan\left(x\right) e^{x}
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(e^{x}\right) + \left(\tan\left(x\right)\right)\left(e^{x}\right)
  5. 5

    Simplify:

    f′(x)=excos⁡(x)2+tan⁡(x)exf'(x) = \frac{e^{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) e^{x}
  6. 6

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

    f′′(x)=excos⁡(x)2−ex2cos⁡(x)−sin⁡(x)(cos⁡(x)2)2+excos⁡(x)2+tan⁡(x)exf''(x) = \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}} + \frac{e^{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) e^{x}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\tan\left(x\right) e^{x}\right] = \frac{e^{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) e^{x} \]
\[ f(x) = \tan\left(x\right) e^{x} \]
\[ u = \tan\left(x\right),\quad v = e^{x} \]
\[ u' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = e^{x} \]
\[ f'(x) = \left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(e^{x}\right) + \left(\tan\left(x\right)\right)\left(e^{x}\right) \]
\[ f'(x) = \frac{e^{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) e^{x} \]
\[ f''(x) = \frac{e^{x} \cos\left(x\right)^{2} - e^{x} 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} + \frac{e^{x}}{\cos\left(x\right)^{2}} + \tan\left(x\right) e^{x} \]

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

xf(x)f'(x)
-20.295712991.0771929
-1-0.572938280.68723967
0.50.900700543.0414768
14.233473113.544999
2-16.14538226.521978
3-2.863123917.630541

Questions

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

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

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).

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

The product rule.

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

13.544999.

Related derivatives

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