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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

\int \frac{8}{( 1 + 4 t ) ^{2}} d t

\frac{d}{d x} (8 + x)

f (x) = ln (x), at x = 6

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 9 x^{3}))

\frac{( 1 - e ^{- s} )}{s ^{2}}

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