What is the derivative of (cos(6x)/(e^x)) ?

The derivative of (cos(6x)/(e^x)) is -(6sin(6x)+cos(6x))/(e^x)

What is the first derivative of (cos(6x)/(e^x)) ?

The first derivative of (cos(6x)/(e^x)) is -(6sin(6x)+cos(6x))/(e^x)

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\frac{d}{d x} (\frac{cos ( 6 x )}{e ^{x}})

- \frac{6 sin ( 6 x ) + cos ( 6 x )}{e ^{x}}

Solution steps

\frac{d}{d x} (\frac{cos ( 6 x )}{e ^{x}})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (cos (6 x)) = - sin (6 x) \cdot 6

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

= \frac{( - sin ( 6 x ) \cdot 6 ) e ^{x} - e ^{x} cos ( 6 x )}{( e ^{x} ) ^{2}}

Simplify

\frac{( - sin ( 6 x ) \cdot 6 ) e ^{x} - e ^{x} cos ( 6 x )}{( e ^{x} ) ^{2}} : - \frac{6 sin ( 6 x ) + cos ( 6 x )}{e ^{x}}

= - \frac{6 sin ( 6 x ) + cos ( 6 x )}{e ^{x}}

y = arctan (x)

x^{2} y^{^{'}} = 2 x^{2} + x y + 2 y^{2}

\int \frac{( 5 - 4 x ^{2} )}{( 3 + 2 x )} d x

\frac{\partial}{\partial x} (x y + x + y)

(y^{- 1})^{^{'}}

What is the derivative of (cos(6x)/(e^x)) ?
The derivative of (cos(6x)/(e^x)) is -(6sin(6x)+cos(6x))/(e^x)
What is the first derivative of (cos(6x)/(e^x)) ?
The first derivative of (cos(6x)/(e^x)) is -(6sin(6x)+cos(6x))/(e^x)