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

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

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

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

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derivative of

y = (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

y^{^{''}} + 3 y^{^{'}} + 0 = 5

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

f (x) = x^{3} - 16 x + 5, at x = 4

\int_{0}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

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

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