What is the derivative of sin(2xdx) ?

The derivative of sin(2xdx) is d(2xcos(2x)+sin(2x))

What is the first derivative of sin(2xdx) ?

The first derivative of sin(2xdx) is d(2xcos(2x)+sin(2x))

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\frac{d}{d x} (sin (2 x) d x)

d (2 x cos (2 x) + sin (2 x))

Solution steps

\frac{d}{d x} (sin (2 x) d x)

Treat

d

as a constant

Take the constant out:

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

= d \frac{d}{d x} (sin (2 x) x)

Apply the Product Rule:

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

f = sin (2 x), g = x

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

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

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

= d (cos (2 x) \cdot 2 x + 1 \cdot sin (2 x))

Simplify

= d (2 x cos (2 x) + sin (2 x))

(x y sin (x) + 2 y cos (x)) d x + (2 x cos (x)) d y = 0

h \to 0 lim (\frac{( 2 - h ) ^{3} - 8}{h})

\int sin (y^{3}) d y

\frac{\partial}{\partial x} (sin (π (3 x - 5 y)))

\int \frac{t ^{3}}{t ^{4} + 9} d t

What is the derivative of sin(2xdx) ?
The derivative of sin(2xdx) is d(2xcos(2x)+sin(2x))
What is the first derivative of sin(2xdx) ?
The first derivative of sin(2xdx) is d(2xcos(2x)+sin(2x))