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

The derivative of (5x-4(2x^3-x^2+1)) is 40x^3-39x^2+8x+5

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

The first derivative of (5x-4(2x^3-x^2+1)) is 40x^3-39x^2+8x+5

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

40 x^{3} - 39 x^{2} + 8 x + 5

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (5 x - 4) = 5

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

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

Simplify

5 (2 x^{3} - x^{2} + 1) + (6 x^{2} - 2 x) (5 x - 4) : 40 x^{3} - 39 x^{2} + 8 x + 5

= 40 x^{3} - 39 x^{2} + 8 x + 5

f (x) = 9 e^{x} + x, (0, 9)

\int \frac{x ^{4}}{2} d x

\frac{d}{d x} (\frac{x sin ( x )}{1 + x})

\int_{0}^{8} \frac{1}{x ^{2} - 6 x + 5} d x

\frac{d}{d x} (sec (5 x^{3} e^{x}))

What is the derivative of (5x-4(2x^3-x^2+1)) ?
The derivative of (5x-4(2x^3-x^2+1)) is 40x^3-39x^2+8x+5
What is the first derivative of (5x-4(2x^3-x^2+1)) ?
The first derivative of (5x-4(2x^3-x^2+1)) is 40x^3-39x^2+8x+5