What is the derivative of (sin(3x))/(tan(4x)) ?

The derivative of (sin(3x))/(tan(4x)) is 3cos(3x)cot(4x)-4csc^2(4x)sin(3x)

What is the first derivative of (sin(3x))/(tan(4x)) ?

The first derivative of (sin(3x))/(tan(4x)) is 3cos(3x)cot(4x)-4csc^2(4x)sin(3x)

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derivative of

\frac{sin ( 3 x )}{tan ( 4 x )}

3 cos (3 x) cot (4 x) - 4 csc^{2} (4 x) sin (3 x)

Solution steps

\frac{d}{d x} (\frac{sin ( 3 x )}{tan ( 4 x )})

Simplify

\frac{sin ( 3 x )}{tan ( 4 x )} : sin (3 x) cot (4 x)

= \frac{d}{d x} (sin (3 x) cot (4 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin (3 x), g = cot (4 x)

= \frac{d}{d x} (sin (3 x)) cot (4 x) + \frac{d}{d x} (cot (4 x)) sin (3 x)

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

\frac{d}{d x} (cot (4 x)) = - csc^{2} (4 x) \cdot 4

= cos (3 x) \cdot 3 cot (4 x) + (- csc^{2} (4 x) \cdot 4) sin (3 x)

Simplify

= 3 cos (3 x) cot (4 x) - 4 csc^{2} (4 x) sin (3 x)

\frac{1}{x - 1}, 21

f (x) = e^{x} (tan (x) - x)

\int_{0}^{\frac{1}{6}} \frac{3}{1 - 9 x ^{2}} d x

\frac{d x}{d t} = x t

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 8, y^{^{'}} (0) = 2

What is the derivative of (sin(3x))/(tan(4x)) ?
The derivative of (sin(3x))/(tan(4x)) is 3cos(3x)cot(4x)-4csc^2(4x)sin(3x)
What is the first derivative of (sin(3x))/(tan(4x)) ?
The first derivative of (sin(3x))/(tan(4x)) is 3cos(3x)cot(4x)-4csc^2(4x)sin(3x)