derivative of 2cos(x+3sin(x)) — step-by-step solution

Let's perform the differentiation:

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

We differentiate using:

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

(where f =

x + 3 \cdot \sin (x)

)

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

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

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

■The derivative of cos(x) is -sin(x):

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

- \frac{d}{d x} (x + 3 \cdot \sin (x)) \cdot \sin (x + 3 \cdot \sin (x))

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

ⓘ The minus sign is placed in front of the product:

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

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

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

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

\frac{d}{d x} (x + 3 \cdot \sin (x))

1 + \frac{d}{d x} (3 \cdot \sin (x))

- 2 \cdot (1 + \frac{d}{d x} (3 \cdot \sin (x))) \cdot \sin (x + 3 \cdot \sin (x))

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

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

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

■The derivative of sin(x) is cos(x):

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

\cos (x)

- 2 \cdot (1 + 3 \cdot \cos (x)) \cdot \sin (x + 3 \cdot \sin (x))

▶Derivative:

- 2 \cdot \sin (x + 3 \cdot \sin (x)) \cdot (3 \cdot \cos (x) + 1)