What is the derivative of 2sin(3x+1) ?

The derivative of 2sin(3x+1) is 6cos(3x)

What is the first derivative of 2sin(3x+1) ?

The first derivative of 2sin(3x+1) is 6cos(3x)

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

6 cos (3 x)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 sin (3 x)) = 6 cos (3 x)

\frac{d}{d x} (1) = 0

= 6 cos (3 x) + 0

Simplify

= 6 cos (3 x)

x \to 0 lim (\frac{e ^{x^{2}} - 1}{x})

\frac{d}{d x} (34 (2 e^{x^{2}} x^{2} + e^{x^{2}}))

y^{^{'}} - 8 y 2 e - 5 x = 0, y (0) = 7

\frac{5}{2 s + 2}

\frac{\partial}{\partial x} (4 e^{2 x})

What is the derivative of 2sin(3x+1) ?
The derivative of 2sin(3x+1) is 6cos(3x)
What is the first derivative of 2sin(3x+1) ?
The first derivative of 2sin(3x+1) is 6cos(3x)