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

The derivative of h(x)=(x^2-3x+2)(2x^3+1) is 10x^4-24x^3+12x^2+2x-3

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

The first derivative of h(x)=(x^2-3x+2)(2x^3+1) is 10x^4-24x^3+12x^2+2x-3

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derivative of

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

10 x^{4} - 24 x^{3} + 12 x^{2} + 2 x - 3

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

(2 x - 3) (2 x^{3} + 1) + 6 x^{2} (x^{2} - 3 x + 2) : 10 x^{4} - 24 x^{3} + 12 x^{2} + 2 x - 3

= 10 x^{4} - 24 x^{3} + 12 x^{2} + 2 x - 3

x^{3}, at x = - 1

\frac{\partial}{\partial x} (6 y x^{5} - 20 y^{2} x^{4})

y = x, y = 2 x

\frac{\partial}{\partial x} (cos (x) sin (y))

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

What is the derivative of h(x)=(x^2-3x+2)(2x^3+1) ?
The derivative of h(x)=(x^2-3x+2)(2x^3+1) is 10x^4-24x^3+12x^2+2x-3
What is the first derivative of h(x)=(x^2-3x+2)(2x^3+1) ?
The first derivative of h(x)=(x^2-3x+2)(2x^3+1) is 10x^4-24x^3+12x^2+2x-3