What is the derivative of sin(2x+pi) ?

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

What is the first derivative of sin(2x+pi) ?

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

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derivative of

sin (2 x + π)

- cos (2 x) \cdot 2

Solution steps

\frac{d}{d x} (sin (2 x + π))

Simplify

sin (2 x + π) : - sin (2 x)

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - cos (2 x) \cdot 2

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

\int 3 x^{2} (x^{3} + 1)^{2} d x

\frac{\partial}{\partial x} (cos (y) + y e^{x})

\int \frac{16}{x ^{2} 16 - x ^{2}} d x

y^{^{'}} = - \frac{y}{( x + \frac{2}{y} )}

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