differentiate sin(2x)/2-x — step-by-step solution

Let's perform the differentiation:

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

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

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

Apply the differentiation rule:

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

with f =

2 x

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of sin(x) is cos(x):

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

\frac{1}{2} \cdot \frac{d}{d x} (2 x) \cdot \cos (2 x)

■The derivative of x is 1:

\frac{d}{d x} (- x)

-1

\frac{1}{2} \cdot \frac{d}{d x} (2 x) \cdot \cos (2 x) - 1

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \frac{d}{d x} (2 x) \cdot \cos (2 x)

\frac{1 \cdot \frac{d}{d x} (2 x) \cdot \cos (2 x)}{2}

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

■A constant factor stays in front of the derivative, (k*u)' = k*u':

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

2

\frac{2 \cdot \cos (2 x)}{2} - 1

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 \cdot \cos (2 x)}{2}

\cos (2 x)

▶Derivative:

\cos (2 x) - 1