What is the derivative of y=(sin(x)+cos(x))/(e^x) ?

The derivative of y=(sin(x)+cos(x))/(e^x) is -(2sin(x))/(e^x)

What is the first derivative of y=(sin(x)+cos(x))/(e^x) ?

The first derivative of y=(sin(x)+cos(x))/(e^x) is -(2sin(x))/(e^x)

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derivative of

y = \frac{sin ( x ) + cos ( x )}{e ^{x}}

- \frac{2 sin ( x )}{e ^{x}}

Solution steps

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{e ^{x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( sin ( x ) + cos ( x ) ) e ^{x} - \frac{d}{d x} ( e ^{x} ) ( sin ( x ) + cos ( x ) )}{( e ^{x} ) ^{2}}

\frac{d}{d x} (sin (x) + cos (x)) = cos (x) - sin (x)

\frac{d}{d x} (e^{x}) = e^{x}

= \frac{( cos ( x ) - sin ( x ) ) e ^{x} - e ^{x} ( sin ( x ) + cos ( x ) )}{( e ^{x} ) ^{2}}

Simplify

\frac{( cos ( x ) - sin ( x ) ) e ^{x} - e ^{x} ( sin ( x ) + cos ( x ) )}{( e ^{x} ) ^{2}} : - \frac{2 sin ( x )}{e ^{x}}

= - \frac{2 sin ( x )}{e ^{x}}

\frac{d}{d x} (- \frac{6 e ^{\frac{6}{x}}}{x ^{2}})

x \cdot cos (x)

\int 6 (36 - 4 x)^{\frac{1}{2}} d x

\int \frac{x ^{3}}{( x ^{2} - 16 )} d x

y = x, y = - x, x = 1

What is the derivative of y=(sin(x)+cos(x))/(e^x) ?
The derivative of y=(sin(x)+cos(x))/(e^x) is -(2sin(x))/(e^x)
What is the first derivative of y=(sin(x)+cos(x))/(e^x) ?
The first derivative of y=(sin(x)+cos(x))/(e^x) is -(2sin(x))/(e^x)