What is the ((4x^3)/3+2x+cos(x)+1)^' ?

The ((4x^3)/3+2x+cos(x)+1)^' is 4x^2-sin(x)+2

What is the first ((4x^3)/3+2x+cos(x)+1)^' ?

The first ((4x^3)/3+2x+cos(x)+1)^' is 4x^2-sin(x)+2

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(\frac{4 x ^{3}}{3} + 2 x + cos (x) + 1)^{^{'}}

4 x^{2} - sin (x) + 2

Solution steps

(\frac{4 x ^{3}}{3} + 2 x + cos (x) + 1)^{'}

Apply the Sum/Difference Rule:

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

= (\frac{4 x ^{3}}{3})^{^{'}} + (2 x)^{^{'}} + (cos (x))^{^{'}} + 1^{^{'}}

(\frac{4 x ^{3}}{3})^{'} = 4 x^{2}

(2 x)^{'} = 2

(cos (x))^{'} = - sin (x)

1^{'} = 0

= 4 x^{2} + 2 - sin (x) + 0

Simplify

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

\int (16 - 4 x^{2}) d x

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

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

\int_{0}^{2} 7 x^{3} x^{2} + 4 d x

\frac{d}{d x} (\frac{x ^{2} + 6 x}{10 - x ^{2}})

What is the ((4x^3)/3+2x+cos(x)+1)^' ?
The ((4x^3)/3+2x+cos(x)+1)^' is 4x^2-sin(x)+2
What is the first ((4x^3)/3+2x+cos(x)+1)^' ?
The first ((4x^3)/3+2x+cos(x)+1)^' is 4x^2-sin(x)+2