What is the derivative of c_{1}cos(2x+c_{2}sin(2x)) ?

The derivative of c_{1}cos(2x+c_{2}sin(2x)) is -2c_{1}sin(2x)+c_{2}cos(2x)*2

What is the first derivative of c_{1}cos(2x+c_{2}sin(2x)) ?

The first derivative of c_{1}cos(2x+c_{2}sin(2x)) is -2c_{1}sin(2x)+c_{2}cos(2x)*2

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\frac{d}{d x} (c_{1} cos (2 x) + c_{2} sin (2 x))

- 2 c_{1} sin (2 x) + c_{2} cos (2 x) \cdot 2

Solution steps

\frac{d}{d x} (c_{1} cos (2 x) + c_{2} sin (2 x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (c_{1} cos (2 x)) + \frac{d}{d x} (c_{2} sin (2 x))

\frac{d}{d x} (c_{1} cos (2 x)) = - 2 c_{1} sin (2 x)

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

= - 2 c_{1} sin (2 x) + c_{2} cos (2 x) \cdot 2

What is the derivative of c_{1}cos(2x+c_{2}sin(2x)) ?
The derivative of c_{1}cos(2x+c_{2}sin(2x)) is -2c_{1}sin(2x)+c_{2}cos(2x)*2
What is the first derivative of c_{1}cos(2x+c_{2}sin(2x)) ?
The first derivative of c_{1}cos(2x+c_{2}sin(2x)) is -2c_{1}sin(2x)+c_{2}cos(2x)*2