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

The derivative of 2x^3(1-x^2^4) is 2(3x^2(1-x^2)^4-8x^4(1-x^2)^3)

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

The first derivative of 2x^3(1-x^2^4) is 2(3x^2(1-x^2)^4-8x^4(1-x^2)^3)

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

2 (3 x^{2} (1 - x^{2})^{4} - 8 x^{4} (1 - x^{2})^{3})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} ((1 - x^{2})^{4}) = - 8 x (1 - x^{2})^{3}

= 2 (3 x^{2} (1 - x^{2})^{4} + (- 8 x (1 - x^{2})^{3}) x^{3})

Simplify

2 (3 x^{2} (1 - x^{2})^{4} + (- 8 x (1 - x^{2})^{3}) x^{3}) : 2 (3 x^{2} (1 - x^{2})^{4} - 8 x^{4} (1 - x^{2})^{3})

= 2 (3 x^{2} (1 - x^{2})^{4} - 8 x^{4} (1 - x^{2})^{3})

x y^{^{'}} - 2 y + 8 = 0

\int e^{4 x + 5} d x

e^{- 8 s}

x \to 2 lim (x^{2} + 4)

\int_{\frac{1}{2}}^{1} x^{- 3} - 8 d x

What is the derivative of 2x^3(1-x^2^4) ?
The derivative of 2x^3(1-x^2^4) is 2(3x^2(1-x^2)^4-8x^4(1-x^2)^3)
What is the first derivative of 2x^3(1-x^2^4) ?
The first derivative of 2x^3(1-x^2^4) is 2(3x^2(1-x^2)^4-8x^4(1-x^2)^3)