What is the derivative of (sin(8x)/(e^{4x)}) ?

The derivative of (sin(8x)/(e^{4x)}) is (4(2cos(8x)-sin(8x)))/(e^{4x)}

What is the first derivative of (sin(8x)/(e^{4x)}) ?

The first derivative of (sin(8x)/(e^{4x)}) is (4(2cos(8x)-sin(8x)))/(e^{4x)}

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\frac{d}{d x} (\frac{sin ( 8 x )}{e ^{4 x}})

\frac{4 ( 2 cos ( 8 x ) - sin ( 8 x ) )}{e ^{4 x}}

Solution steps

\frac{d}{d x} (\frac{sin ( 8 x )}{e ^{4 x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( sin ( 8 x ) ) e ^{4 x} - \frac{d}{d x} ( e ^{4 x} ) sin ( 8 x )}{( e ^{4 x} ) ^{2}}

\frac{d}{d x} (sin (8 x)) = cos (8 x) \cdot 8

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

= \frac{cos ( 8 x ) \cdot 8 e ^{4 x} - e ^{4 x} \cdot 4 sin ( 8 x )}{( e ^{4 x} ) ^{2}}

Simplify

\frac{cos ( 8 x ) \cdot 8 e ^{4 x} - e ^{4 x} \cdot 4 sin ( 8 x )}{( e ^{4 x} ) ^{2}} : \frac{4 ( 2 cos ( 8 x ) - sin ( 8 x ) )}{e ^{4 x}}

= \frac{4 ( 2 cos ( 8 x ) - sin ( 8 x ) )}{e ^{4 x}}

\frac{\partial}{\partial x} (ln (3 y^{2} - x^{3} - 3))

x \to 0 lim (\frac{5 ^{x} - 1 3 ^{x}}{x})

x \to 0 lim (x \cdot sin (\frac{1}{x}))

\frac{\partial}{\partial x} (ln (x) - 4)

\frac{d}{d x} (y^{5} sin (5 x))

What is the derivative of (sin(8x)/(e^{4x)}) ?
The derivative of (sin(8x)/(e^{4x)}) is (4(2cos(8x)-sin(8x)))/(e^{4x)}
What is the first derivative of (sin(8x)/(e^{4x)}) ?
The first derivative of (sin(8x)/(e^{4x)}) is (4(2cos(8x)-sin(8x)))/(e^{4x)}