What is the derivative of (sin(6x)/(e^{3x)}) ?

The derivative of (sin(6x)/(e^{3x)}) is (3(2cos(6x)-sin(6x)))/(e^{3x)}

What is the first derivative of (sin(6x)/(e^{3x)}) ?

The first derivative of (sin(6x)/(e^{3x)}) is (3(2cos(6x)-sin(6x)))/(e^{3x)}

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\frac{d}{d x} (\frac{sin ( 6 x )}{e ^{3 x}})

\frac{3 ( 2 cos ( 6 x ) - sin ( 6 x ) )}{e ^{3 x}}

Solution steps

\frac{d}{d x} (\frac{sin ( 6 x )}{e ^{3 x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( sin ( 6 x ) ) e ^{3 x} - \frac{d}{d x} ( e ^{3 x} ) sin ( 6 x )}{( e ^{3 x} ) ^{2}}

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

\frac{d}{d x} (e^{3 x}) = e^{3 x} \cdot 3

= \frac{cos ( 6 x ) \cdot 6 e ^{3 x} - e ^{3 x} \cdot 3 sin ( 6 x )}{( e ^{3 x} ) ^{2}}

Simplify

\frac{cos ( 6 x ) \cdot 6 e ^{3 x} - e ^{3 x} \cdot 3 sin ( 6 x )}{( e ^{3 x} ) ^{2}} : \frac{3 ( 2 cos ( 6 x ) - sin ( 6 x ) )}{e ^{3 x}}

= \frac{3 ( 2 cos ( 6 x ) - sin ( 6 x ) )}{e ^{3 x}}

d er i v a t i v e f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 5)

\frac{1}{3} \frac{d x}{d t} - 2 e^{- t} = - x

\int_{l n (8)}^{l n (64)} e^{\frac{x}{3}} d x

x \to 0 + lim (x - x)

\int x + \frac{1}{2 x} d x

What is the derivative of (sin(6x)/(e^{3x)}) ?
The derivative of (sin(6x)/(e^{3x)}) is (3(2cos(6x)-sin(6x)))/(e^{3x)}
What is the first derivative of (sin(6x)/(e^{3x)}) ?
The first derivative of (sin(6x)/(e^{3x)}) is (3(2cos(6x)-sin(6x)))/(e^{3x)}