What is the derivative of-0.8sin(x+cos(2x)) ?

The derivative of-0.8sin(x+cos(2x)) is -0.8cos(x)-sin(2x)*2

What is the first derivative of-0.8sin(x+cos(2x)) ?

The first derivative of-0.8sin(x+cos(2x)) is -0.8cos(x)-sin(2x)*2

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\frac{d}{d x} (- 0.8 sin (x) + cos (2 x))

- 0.8 cos (x) - sin (2 x) \cdot 2

Solution steps

\frac{d}{d x} (- 0.8 sin (x) + cos (2 x))

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (0.8 sin (x)) + \frac{d}{d x} (cos (2 x))

\frac{d}{d x} (0.8 sin (x)) = 0.8 cos (x)

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

= - 0.8 cos (x) - sin (2 x) \cdot 2

y^{^{'}} = y (x y^{5} + 3)

x \to 8 lim (500 x - 6 x^{2})

x \to 4 - lim (\frac{x}{x - 4})

x \to - 3 lim (x^{3} - 2 x^{2} + x + 7)

y = 3 ln (ln (x))

What is the derivative of-0.8sin(x+cos(2x)) ?
The derivative of-0.8sin(x+cos(2x)) is -0.8cos(x)-sin(2x)*2
What is the first derivative of-0.8sin(x+cos(2x)) ?
The first derivative of-0.8sin(x+cos(2x)) is -0.8cos(x)-sin(2x)*2