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Derivative of excos⁡(x)\frac{e^{x}}{\cos\left(x\right)}

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

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

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Graph of f(x) = exp(x) / cos(x) and its derivative-3-2-1123-60-40-2020406080slope 12.87f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x) / cos(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 12.866.

How to work it out

  1. 1

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

    f(x)=excos⁡(x)f(x) = \frac{e^{x}}{\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=exu = e^{x} and v=cos⁡(x)v = \cos\left(x\right).

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

    Differentiate each piece:

    u′=ex,v′=−sin⁡(x)u' = e^{x},\quad v' = -\sin\left(x\right)
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=excos⁡(x)−ex−sin⁡(x)cos⁡(x)2f'(x) = \frac{e^{x} \cos\left(x\right) - e^{x} -\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)=(excos⁡(x)+ex−sin⁡(x)−(ex−sin⁡(x)+ex−cos⁡(x)))cos⁡(x)2−(excos⁡(x)−ex−sin⁡(x))2cos⁡(x)−sin⁡(x)(cos⁡(x)2)2f''(x) = \frac{\left(e^{x} \cos\left(x\right) + e^{x} -\sin\left(x\right) - \left(e^{x} -\sin\left(x\right) + e^{x} -\cos\left(x\right)\right)\right) \cos\left(x\right)^{2} - \left(e^{x} \cos\left(x\right) - e^{x} -\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
\[ \frac{d}{dx}\left[\frac{e^{x}}{\cos\left(x\right)}\right] = \frac{e^{x} \cos\left(x\right) - e^{x} -\sin\left(x\right)}{\cos\left(x\right)^{2}} \]
\[ f(x) = \frac{e^{x}}{\cos\left(x\right)} \]
\[ u = e^{x},\quad v = \cos\left(x\right) \]
\[ u' = e^{x},\quad v' = -\sin\left(x\right) \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(\cos\left(x\right)\right) - \left(e^{x}\right)\left(-\sin\left(x\right)\right)}{\left(\cos\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{e^{x} \cos\left(x\right) - e^{x} -\sin\left(x\right)}{\cos\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(e^{x} \cos\left(x\right) + e^{x} -\sin\left(x\right) - \left(e^{x} -\sin\left(x\right) + e^{x} -\cos\left(x\right)\right)\right) \cos\left(x\right)^{2} - \left(e^{x} \cos\left(x\right) - e^{x} -\sin\left(x\right)\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-0.32521041-1.0358081
-10.68087705-0.37952613
0.51.87870792.9050507
15.031038712.866417
2-17.75588721.041434
3-20.288575-17.396509

Questions

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

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

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

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

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

The quotient rule.

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

12.866417.

Related derivatives

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