What is the (12x+x^2+0.05x^3)^' ?

The (12x+x^2+0.05x^3)^' is 12+2x+0.15x^2

What is the first (12x+x^2+0.05x^3)^' ?

The first (12x+x^2+0.05x^3)^' is 12+2x+0.15x^2

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(12 x + x^{2} + 0.05 x^{3})^{^{'}}

12 + 2 x + 0.15 x^{2}

Solution steps

(12 x + x^{2} + 0.05 x^{3})^{'}

Apply the Sum/Difference Rule:

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

= (12 x)^{^{'}} + (x^{2})^{^{'}} + (0.05 x^{3})^{^{'}}

(12 x)^{'} = 12

(x^{2})^{'} = 2 x

(0.05 x^{3})^{'} = 0.15 x^{2}

= 12 + 2 x + 0.15 x^{2}

\int_{- 1}^{1} e^{y} - (y^{2} - 2) d y

n = 2 \sum \infty \frac{3}{4 ^{n}}

\frac{\partial}{\partial x} (cos^{7} (x^{4} y^{6}))

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2}))

y^{^{''}} + 10 y^{^{'}} + 25 y = 4 e^{- 5 t}

What is the (12x+x^2+0.05x^3)^' ?
The (12x+x^2+0.05x^3)^' is 12+2x+0.15x^2
What is the first (12x+x^2+0.05x^3)^' ?
The first (12x+x^2+0.05x^3)^' is 12+2x+0.15x^2