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

The derivative of 4(x^3-x^4) is 16(x^3-x)^3(3x^2-1)

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

The first derivative of 4(x^3-x^4) is 16(x^3-x)^3(3x^2-1)

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

16 (x^{3} - x)^{3} (3 x^{2} - 1)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 4 \cdot 4 (x^{3} - x)^{3} (3 x^{2} - 1)

Simplify

= 16 (x^{3} - x)^{3} (3 x^{2} - 1)

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

3 x^{2} y^{2} d x + y d y = 5 x y

\int_{1}^{9} \frac{1}{x} d x

\int (x^{2} - 4) d x

f (x) = x^{3} - 2 x + 2, (3, 23)

What is the derivative of 4(x^3-x^4) ?
The derivative of 4(x^3-x^4) is 16(x^3-x)^3(3x^2-1)
What is the first derivative of 4(x^3-x^4) ?
The first derivative of 4(x^3-x^4) is 16(x^3-x)^3(3x^2-1)