What is the derivative of f(2)=sin(3x)-4x^4+2x^3+3 ?

The derivative of f(2)=sin(3x)-4x^4+2x^3+3 is -16x^3+6x^2+3cos(3x)

What is the first derivative of f(2)=sin(3x)-4x^4+2x^3+3 ?

The first derivative of f(2)=sin(3x)-4x^4+2x^3+3 is -16x^3+6x^2+3cos(3x)

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derivative of

f (2) = sin (3 x) - 4 x^{4} + 2 x^{3} + 3

- 16 x^{3} + 6 x^{2} + 3 cos (3 x)

Solution steps

\frac{d}{d x} (sin (3 x) - 4 x^{4} + 2 x^{3} + 3)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (sin (3 x)) - \frac{d}{d x} (4 x^{4}) + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (3)

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

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

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

\frac{d}{d x} (3) = 0

= cos (3 x) \cdot 3 - 16 x^{3} + 6 x^{2} + 0

Simplify

= - 16 x^{3} + 6 x^{2} + 3 cos (3 x)

\int cos (x) (2 + 4 sin^{2} (x)) d x

f (x) = \frac{2 x}{1 + x ^{2}}, at x = (4)

x y^{^{'}} - 2 y = x^{4}, y (1) = 0

\frac{d}{d x} ((4 x - 8) e^{4 x})

What is the derivative of f(2)=sin(3x)-4x^4+2x^3+3 ?
The derivative of f(2)=sin(3x)-4x^4+2x^3+3 is -16x^3+6x^2+3cos(3x)
What is the first derivative of f(2)=sin(3x)-4x^4+2x^3+3 ?
The first derivative of f(2)=sin(3x)-4x^4+2x^3+3 is -16x^3+6x^2+3cos(3x)