What is the derivative of (sin(3x)/(7x)) ?

The derivative of (sin(3x)/(7x)) is (3xcos(3x)-sin(3x))/(7x^2)

What is the first derivative of (sin(3x)/(7x)) ?

The first derivative of (sin(3x)/(7x)) is (3xcos(3x)-sin(3x))/(7x^2)

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\frac{d}{d x} (\frac{sin ( 3 x )}{7 x})

\frac{3 x cos ( 3 x ) - sin ( 3 x )}{7 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{sin ( 3 x )}{7 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{7} \frac{d}{d x} (\frac{sin ( 3 x )}{x})

Apply the Quotient Rule:

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

= \frac{1}{7} \cdot \frac{\frac{d}{d x} ( sin ( 3 x ) ) x - \frac{d x}{d x} sin ( 3 x )}{x ^{2}}

\frac{d}{d x} (sin (3 x)) = cos (3 x) \cdot 3

\frac{d x}{d x} = 1

= \frac{1}{7} \cdot \frac{cos ( 3 x ) \cdot 3 x - 1 \cdot sin ( 3 x )}{x ^{2}}

Simplify

\frac{1}{7} \cdot \frac{cos ( 3 x ) \cdot 3 x - 1 \cdot sin ( 3 x )}{x ^{2}} : \frac{3 x cos ( 3 x ) - sin ( 3 x )}{7 x ^{2}}

= \frac{3 x cos ( 3 x ) - sin ( 3 x )}{7 x ^{2}}

\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{1}{x})

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

\int sec^{6} (4 x) tan (4 x) d x

a re a f (x) = x^{2}, 0, 4

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (1) = e^{2}, y^{^{'}} (1) = 3 e^{2}

What is the derivative of (sin(3x)/(7x)) ?
The derivative of (sin(3x)/(7x)) is (3xcos(3x)-sin(3x))/(7x^2)
What is the first derivative of (sin(3x)/(7x)) ?
The first derivative of (sin(3x)/(7x)) is (3xcos(3x)-sin(3x))/(7x^2)