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

The derivative of (3x^4(4x^3+4x)) is 3(28x^6+20x^4)

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

The first derivative of (3x^4(4x^3+4x)) is 3(28x^6+20x^4)

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

3 (28 x^{6} + 20 x^{4})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{4}, g = 4 x^{3} + 4 x

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

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

\frac{d}{d x} (4 x^{3} + 4 x) = 12 x^{2} + 4

= 3 (4 x^{3} (4 x^{3} + 4 x) + (12 x^{2} + 4) x^{4})

Expand

4 x^{3} (4 x^{3} + 4 x) + (12 x^{2} + 4) x^{4} : 28 x^{6} + 20 x^{4}

= 3 (28 x^{6} + 20 x^{4})

\int \frac{e ^{2 x}}{e ^{x} - 5} d x

f (x) = (7 x^{2} - 12 x) e^{x}

3 x^{2}, x^{2} + 6

\frac{d y}{d x} = 6 x y

2 e^{x}

What is the derivative of (3x^4(4x^3+4x)) ?
The derivative of (3x^4(4x^3+4x)) is 3(28x^6+20x^4)
What is the first derivative of (3x^4(4x^3+4x)) ?
The first derivative of (3x^4(4x^3+4x)) is 3(28x^6+20x^4)