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The derivative of the function/ e^(-x)(cosx+sinx)

Derivative of e^(-x)(cosx+sinx)

The solution

You have entered [src]

 -x                  
E  *(cos(x) + sin(x))

$$e^{- x} \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right)$$

E^(-x)*(cos(x) + sin(x))

Detail solution

Apply the quotient rule, which is:

$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$

$f{\left(x \right)} = \sin{\left(x \right)} + \cos{\left(x \right)}$ and $g{\left(x \right)} = e^{x}$ .

To find $\frac{d}{d x} f{\left(x \right)}$ :
1. Differentiate $\sin{\left(x \right)} + \cos{\left(x \right)}$ term by term:
  1. The derivative of cosine is negative sine:
    
    $\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$
  2. The derivative of sine is cosine:
    
    $\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$
  The result is: $- \sin{\left(x \right)} + \cos{\left(x \right)}$
To find $\frac{d}{d x} g{\left(x \right)}$ :
1. The derivative of $e^{x}$ is itself.
Now plug in to the quotient rule:

$\left(\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{x} - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{x}\right) e^{- 2 x}$
Now simplify:

$- 2 e^{- x} \sin{\left(x \right)}$

The answer is:

$- 2 e^{- x} \sin{\left(x \right)}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

                    -x                      -x
(-sin(x) + cos(x))*e   - (cos(x) + sin(x))*e

$$\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x}$$

Simplify

The second derivative [src]

                      -x
2*(-cos(x) + sin(x))*e

$$2 \left(\sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{- x}$$

Simplify

The third derivative [src]

          -x
4*cos(x)*e

$$4 e^{- x} \cos{\left(x \right)}$$

Simplify