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

The derivative of 2x^3+4x^2-3x is 6x^2+8x-3

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

The first derivative of 2x^3+4x^2-3x is 6x^2+8x-3

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

6 x^{2} + 8 x - 3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (4 x^{2}) = 8 x

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

= 6 x^{2} + 8 x - 3

\int_{1}^{\infty} \frac{1}{x ^{4}} d x

\int e^{(x + 1)^{2}} (x + 1) d x

f (x) = \frac{1}{x} - 2, (- 1, - 3)

g (x) = x^{3} \cdot e^{x}

\int_{0}^{\frac{π}{6}} (3 sec^{2} (x)) d x

What is the derivative of 2x^3+4x^2-3x ?
The derivative of 2x^3+4x^2-3x is 6x^2+8x-3
What is the first derivative of 2x^3+4x^2-3x ?
The first derivative of 2x^3+4x^2-3x is 6x^2+8x-3