What is the derivative of 3x^4-2x^3-12x^2+18x+15 ?

The derivative of 3x^4-2x^3-12x^2+18x+15 is 12x^3-6x^2-24x+18

What is the first derivative of 3x^4-2x^3-12x^2+18x+15 ?

The first derivative of 3x^4-2x^3-12x^2+18x+15 is 12x^3-6x^2-24x+18

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\frac{d}{d x} (3 x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 15)

12 x^{3} - 6 x^{2} - 24 x + 18

Solution steps

\frac{d}{d x} (3 x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 15)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{4}) - \frac{d}{d x} (2 x^{3}) - \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (18 x) + \frac{d}{d x} (15)

\frac{d}{d x} (3 x^{4}) = 12 x^{3}

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

\frac{d}{d x} (12 x^{2}) = 24 x

\frac{d}{d x} (18 x) = 18

\frac{d}{d x} (15) = 0

= 12 x^{3} - 6 x^{2} - 24 x + 18 + 0

Simplify

= 12 x^{3} - 6 x^{2} - 24 x + 18

g (x) = x + x

2 sin (x), 3 cos (x)

\frac{d}{d x} (\frac{1}{ln ( x ^{2} )})

\frac{\partial}{\partial y} (x^{2} - y^{2})

\frac{\partial}{\partial θ} (ρ sin (φ) cos (θ))

What is the derivative of 3x^4-2x^3-12x^2+18x+15 ?
The derivative of 3x^4-2x^3-12x^2+18x+15 is 12x^3-6x^2-24x+18
What is the first derivative of 3x^4-2x^3-12x^2+18x+15 ?
The first derivative of 3x^4-2x^3-12x^2+18x+15 is 12x^3-6x^2-24x+18