WordGPT

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

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

f(x) =
\frac{d}{dx}\left[\frac{\cos\left(x\right)}{e^{x}}\right] = \frac{-\left(\sin\left(x\right) e^{x}\right) - \cos\left(x\right) e^{x}}{e^{x}^{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) = cos(x) / exp(x) and its derivative-3-2-1123-10-55101520slope -0.5083f(x)f'(x)tangent at x = 1
Graph of f(x) = cos(x) / exp(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.50833.

How to work it out

  1. 1

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

    f(x)=cos⁡(x)exf(x) = \frac{\cos\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=cos⁡(x)u = \cos\left(x\right) and v=exv = e^{x}.

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

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

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

xf(x)f'(x)
-2-3.07493239.793782
-11.46869390.81866135
0.50.53228073-0.82306702
10.19876611-0.50832599
2-0.05631935-0.066740675
3-0.0492888240.042262873

Questions

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

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

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

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

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

The quotient rule.

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

-0.50832599.

Related derivatives

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