What is the derivative of f(x)=8(x^3-x)^4 ?

The derivative of f(x)=8(x^3-x)^4 is 32(x^3-x)^3(3x^2-1)

What is the first derivative of f(x)=8(x^3-x)^4 ?

The first derivative of f(x)=8(x^3-x)^4 is 32(x^3-x)^3(3x^2-1)

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derivative of

f (x) = 8 (x^{3} - x)^{4}

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

Solution steps

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

Take the constant out:

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

= 8 \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

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

Simplify

= 32 (x^{3} - x)^{3} (3 x^{2} - 1)

y^{^{''}} + 2 y = 10

\frac{( s + 1 )}{s ^{2} + s + 1}

\frac{d y}{d x} = x e^{- y}

\int_{7}^{15} ln (x^{2}) d x

y^{^{'}} = \frac{x ^{8} - 2 y}{x}

What is the derivative of f(x)=8(x^3-x)^4 ?
The derivative of f(x)=8(x^3-x)^4 is 32(x^3-x)^3(3x^2-1)
What is the first derivative of f(x)=8(x^3-x)^4 ?
The first derivative of f(x)=8(x^3-x)^4 is 32(x^3-x)^3(3x^2-1)