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

The derivative of (3x^2+2x(x^4-3x+1)) is 18x^5+10x^4-27x^2-6x+2

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

The first derivative of (3x^2+2x(x^4-3x+1)) is 18x^5+10x^4-27x^2-6x+2

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

18 x^{5} + 10 x^{4} - 27 x^{2} - 6 x + 2

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

(6 x + 2) (x^{4} - 3 x + 1) + (4 x^{3} - 3) (3 x^{2} + 2 x) : 18 x^{5} + 10 x^{4} - 27 x^{2} - 6 x + 2

= 18 x^{5} + 10 x^{4} - 27 x^{2} - 6 x + 2

y^{^{'''}} + y^{^{'}} = sec (x)

\frac{\partial}{\partial x} (x^{2 y})

sec (x) \frac{d y}{d x} = e^{y}

\int cos^{3} (x) (1 + sin (x)) d x

\frac{d y}{d x}, ln (x y) + 9 x = 40

What is the derivative of (3x^2+2x(x^4-3x+1)) ?
The derivative of (3x^2+2x(x^4-3x+1)) is 18x^5+10x^4-27x^2-6x+2
What is the first derivative of (3x^2+2x(x^4-3x+1)) ?
The first derivative of (3x^2+2x(x^4-3x+1)) is 18x^5+10x^4-27x^2-6x+2