What is the derivative of (sin(4x)/(2+cos(4x))) ?

The derivative of (sin(4x)/(2+cos(4x))) is (4cos(4x)(2+cos(4x))+4sin^2(4x))/((2+cos(4x))^2)

What is the first derivative of (sin(4x)/(2+cos(4x))) ?

The first derivative of (sin(4x)/(2+cos(4x))) is (4cos(4x)(2+cos(4x))+4sin^2(4x))/((2+cos(4x))^2)

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\frac{d}{d x} (\frac{sin ( 4 x )}{2 + cos ( 4 x )})

\frac{4 cos ( 4 x ) ( 2 + cos ( 4 x ) ) + 4 sin ^{2} ( 4 x )}{( 2 + cos ( 4 x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{sin ( 4 x )}{2 + cos ( 4 x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( sin ( 4 x ) ) ( 2 + cos ( 4 x ) ) - \frac{d}{d x} ( 2 + cos ( 4 x ) ) sin ( 4 x )}{( 2 + cos ( 4 x ) ) ^{2}}

\frac{d}{d x} (sin (4 x)) = cos (4 x) \cdot 4

\frac{d}{d x} (2 + cos (4 x)) = - 4 sin (4 x)

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

Simplify

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

= \frac{4 cos ( 4 x ) ( 2 + cos ( 4 x ) ) + 4 sin ^{2} ( 4 x )}{( 2 + cos ( 4 x ) ) ^{2}}

What is the derivative of (sin(4x)/(2+cos(4x))) ?
The derivative of (sin(4x)/(2+cos(4x))) is (4cos(4x)(2+cos(4x))+4sin^2(4x))/((2+cos(4x))^2)
What is the first derivative of (sin(4x)/(2+cos(4x))) ?
The first derivative of (sin(4x)/(2+cos(4x))) is (4cos(4x)(2+cos(4x))+4sin^2(4x))/((2+cos(4x))^2)