What is the derivative of (2cos(2x+sin(3x)+3)^2) ?

The derivative of (2cos(2x+sin(3x)+3)^2) is 2(2cos(2x)+sin(3x)+3)(3cos(3x)-4sin(2x))

What is the first derivative of (2cos(2x+sin(3x)+3)^2) ?

The first derivative of (2cos(2x+sin(3x)+3)^2) is 2(2cos(2x)+sin(3x)+3)(3cos(3x)-4sin(2x))

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\frac{d}{d x} ((2 cos (2 x) + sin (3 x) + 3)^{2})

2 (2 cos (2 x) + sin (3 x) + 3) (3 cos (3 x) - 4 sin (2 x))

Solution steps

\frac{d}{d x} ((2 cos (2 x) + sin (3 x) + 3)^{2})

Apply the chain rule:

2 (2 cos (2 x) + sin (3 x) + 3) \frac{d}{d x} (2 cos (2 x) + sin (3 x) + 3)

= 2 (2 cos (2 x) + sin (3 x) + 3) \frac{d}{d x} (2 cos (2 x) + sin (3 x) + 3)

\frac{d}{d x} (2 cos (2 x) + sin (3 x) + 3) = 3 cos (3 x) - 4 sin (2 x)

= 2 (2 cos (2 x) + sin (3 x) + 3) (3 cos (3 x) - 4 sin (2 x))

y = \frac{A x ^{2}}{x ^{2} - 1}

\frac{d y}{d t} - 2 t y = - 9 t^{2} e^{t^{2}}

\frac{d y}{d x} = 2 y^{2} + x y^{2}

\int \frac{1}{x ^{2} 2 + 9 x ^{2}} d x

What is the derivative of (2cos(2x+sin(3x)+3)^2) ?
The derivative of (2cos(2x+sin(3x)+3)^2) is 2(2cos(2x)+sin(3x)+3)(3cos(3x)-4sin(2x))
What is the first derivative of (2cos(2x+sin(3x)+3)^2) ?
The first derivative of (2cos(2x+sin(3x)+3)^2) is 2(2cos(2x)+sin(3x)+3)(3cos(3x)-4sin(2x))