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

The derivative of f(x)=(5x-5)(2x^3-x^2+1) is 40x^3-45x^2+10x+5

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

The first derivative of f(x)=(5x-5)(2x^3-x^2+1) is 40x^3-45x^2+10x+5

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derivative of

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

40 x^{3} - 45 x^{2} + 10 x + 5

Solution steps

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

Apply the Product Rule:

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

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

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

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

Simplify

5 (2 x^{3} - x^{2} + 1) + (6 x^{2} - 2 x) (5 x - 5) : 40 x^{3} - 45 x^{2} + 10 x + 5

= 40 x^{3} - 45 x^{2} + 10 x + 5

\frac{\partial}{\partial u} (u^{5} + v^{9})

\int_{1}^{e^{4}} \frac{ln ^{3} ( x ^{2} )}{x} d x

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

\frac{\partial}{\partial x} (sin (x^{4} + y^{3}))

f (x) = x^{2} - 3, (4, 13)

What is the derivative of f(x)=(5x-5)(2x^3-x^2+1) ?
The derivative of f(x)=(5x-5)(2x^3-x^2+1) is 40x^3-45x^2+10x+5
What is the first derivative of f(x)=(5x-5)(2x^3-x^2+1) ?
The first derivative of f(x)=(5x-5)(2x^3-x^2+1) is 40x^3-45x^2+10x+5