What is the derivative of 6x^2(4x^5-3x+2) ?

The derivative of 6x^2(4x^5-3x+2) is 6(28x^6-9x^2+4x)

What is the first derivative of 6x^2(4x^5-3x+2) ?

The first derivative of 6x^2(4x^5-3x+2) is 6(28x^6-9x^2+4x)

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\frac{d}{d x} (6 x^{2} (4 x^{5} - 3 x + 2))

6 (28 x^{6} - 9 x^{2} + 4 x)

Solution steps

\frac{d}{d x} (6 x^{2} (4 x^{5} - 3 x + 2))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} (x^{2} (4 x^{5} - 3 x + 2))

Apply the Product Rule:

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

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

= 6 (\frac{d}{d x} (x^{2}) (4 x^{5} - 3 x + 2) + \frac{d}{d x} (4 x^{5} - 3 x + 2) x^{2})

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (4 x^{5} - 3 x + 2) = 20 x^{4} - 3

= 6 (2 x (4 x^{5} - 3 x + 2) + (20 x^{4} - 3) x^{2})

Expand

2 x (4 x^{5} - 3 x + 2) + (20 x^{4} - 3) x^{2} : 28 x^{6} - 9 x^{2} + 4 x

= 6 (28 x^{6} - 9 x^{2} + 4 x)

x \to 0 + lim (x^{3} (ln (x))^{2})

y = x (1 + x^{2})^{10}

x \to 0 - lim (\frac{x}{x})

\int (3 y - 4 y^{3})^{3} d y

y = ln (\frac{x}{2}), at x = 4

What is the derivative of 6x^2(4x^5-3x+2) ?
The derivative of 6x^2(4x^5-3x+2) is 6(28x^6-9x^2+4x)
What is the first derivative of 6x^2(4x^5-3x+2) ?
The first derivative of 6x^2(4x^5-3x+2) is 6(28x^6-9x^2+4x)