What is the (9x+0.8x^2+0.07x^3)^' ?

The (9x+0.8x^2+0.07x^3)^' is 9+1.6x+0.21x^2

What is the first (9x+0.8x^2+0.07x^3)^' ?

The first (9x+0.8x^2+0.07x^3)^' is 9+1.6x+0.21x^2

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(9 x + 0.8 x^{2} + 0.07 x^{3})^{^{'}}

9 + 1.6 x + 0.21 x^{2}

Solution steps

(9 x + 0.8 x^{2} + 0.07 x^{3})^{'}

Apply the Sum/Difference Rule:

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

= (9 x)^{^{'}} + (0.8 x^{2})^{^{'}} + (0.07 x^{3})^{^{'}}

(9 x)^{'} = 9

(0.8 x^{2})^{'} = 1.6 x

(0.07 x^{3})^{'} = 0.21 x^{2}

= 9 + 1.6 x + 0.21 x^{2}

y = cos (x)

ln (\frac{1 + x}{1 - x})

\frac{d}{d x} (lo g_{5} (4 x^{2} + 5 x - 8))

\frac{\partial}{\partial x} (x y + \frac{y}{x})

\int \frac{cos ^{5} ( 5 x + 1 )}{sin ^{4} ( 5 x + 1 )} d x

What is the (9x+0.8x^2+0.07x^3)^' ?
The (9x+0.8x^2+0.07x^3)^' is 9+1.6x+0.21x^2
What is the first (9x+0.8x^2+0.07x^3)^' ?
The first (9x+0.8x^2+0.07x^3)^' is 9+1.6x+0.21x^2