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

The derivative of (3x^3+4(x^4-3x)) is 21x^6-20x^3-12

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

The first derivative of (3x^3+4(x^4-3x)) is 21x^6-20x^3-12

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

21 x^{6} - 20 x^{3} - 12

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 x^{3} + 4, g = x^{4} - 3 x

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

\frac{d}{d x} (3 x^{3} + 4) = 9 x^{2}

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

= 9 x^{2} (x^{4} - 3 x) + (4 x^{3} - 3) (3 x^{3} + 4)

Simplify

9 x^{2} (x^{4} - 3 x) + (4 x^{3} - 3) (3 x^{3} + 4) : 21 x^{6} - 20 x^{3} - 12

= 21 x^{6} - 20 x^{3} - 12

f (x) = \frac{7}{x}, at x = 2

\int_{1}^{18} \frac{3}{z} d z

- x + 7 x^{2} - 15 x

y^{^{'}} = 7 y^{2} + x y^{2}, y (0) = 1

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 18} d x

What is the derivative of (3x^3+4(x^4-3x)) ?
The derivative of (3x^3+4(x^4-3x)) is 21x^6-20x^3-12
What is the first derivative of (3x^3+4(x^4-3x)) ?
The first derivative of (3x^3+4(x^4-3x)) is 21x^6-20x^3-12