What is the derivative of (2x^4-1)(5x^3+6x) ?

The derivative of (2x^4-1)(5x^3+6x) is 70x^6+60x^4-15x^2-6

What is the first derivative of (2x^4-1)(5x^3+6x) ?

The first derivative of (2x^4-1)(5x^3+6x) is 70x^6+60x^4-15x^2-6

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derivative of

(2 x^{4} - 1) (5 x^{3} + 6 x)

70 x^{6} + 60 x^{4} - 15 x^{2} - 6

Solution steps

\frac{d}{d x} ((2 x^{4} - 1) (5 x^{3} + 6 x))

Apply the Product Rule:

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

f = 2 x^{4} - 1, g = 5 x^{3} + 6 x

= \frac{d}{d x} (2 x^{4} - 1) (5 x^{3} + 6 x) + \frac{d}{d x} (5 x^{3} + 6 x) (2 x^{4} - 1)

\frac{d}{d x} (2 x^{4} - 1) = 8 x^{3}

\frac{d}{d x} (5 x^{3} + 6 x) = 15 x^{2} + 6

= 8 x^{3} (5 x^{3} + 6 x) + (15 x^{2} + 6) (2 x^{4} - 1)

Simplify

8 x^{3} (5 x^{3} + 6 x) + (15 x^{2} + 6) (2 x^{4} - 1) : 70 x^{6} + 60 x^{4} - 15 x^{2} - 6

= 70 x^{6} + 60 x^{4} - 15 x^{2} - 6

\frac{5}{x}

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

\int_{0}^{1} x^{7} x^{4} + 1 d x

\frac{d y}{d x} = \frac{e ^{x} - 1}{e ^{y}}

\int_{1}^{4} \frac{x ^{2} - 1}{x} d x

What is the derivative of (2x^4-1)(5x^3+6x) ?
The derivative of (2x^4-1)(5x^3+6x) is 70x^6+60x^4-15x^2-6
What is the first derivative of (2x^4-1)(5x^3+6x) ?
The first derivative of (2x^4-1)(5x^3+6x) is 70x^6+60x^4-15x^2-6