What is the derivative of y=(4x^3+5)(x^4-3x) ?

The derivative of y=(4x^3+5)(x^4-3x) is 28x^6-28x^3-15

What is the first derivative of y=(4x^3+5)(x^4-3x) ?

The first derivative of y=(4x^3+5)(x^4-3x) is 28x^6-28x^3-15

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derivative of

y = (4 x^{3} + 5) (x^{4} - 3 x)

28 x^{6} - 28 x^{3} - 15

Solution steps

\frac{d}{d x} ((4 x^{3} + 5) (x^{4} - 3 x))

Apply the Product Rule:

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

f = 4 x^{3} + 5, g = x^{4} - 3 x

= \frac{d}{d x} (4 x^{3} + 5) (x^{4} - 3 x) + \frac{d}{d x} (x^{4} - 3 x) (4 x^{3} + 5)

\frac{d}{d x} (4 x^{3} + 5) = 12 x^{2}

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

= 12 x^{2} (x^{4} - 3 x) + (4 x^{3} - 3) (4 x^{3} + 5)

Simplify

12 x^{2} (x^{4} - 3 x) + (4 x^{3} - 3) (4 x^{3} + 5) : 28 x^{6} - 28 x^{3} - 15

= 28 x^{6} - 28 x^{3} - 15

n = 1 \sum \infty n + \frac{1}{n ^{2}}

\frac{d y}{d x} = 2 x

\int_{1}^{5} π (x - 1) d x

n = 1 \sum \infty n 7

\int \frac{( y + 1 ) ^{2}}{y} d y

What is the derivative of y=(4x^3+5)(x^4-3x) ?
The derivative of y=(4x^3+5)(x^4-3x) is 28x^6-28x^3-15
What is the first derivative of y=(4x^3+5)(x^4-3x) ?
The first derivative of y=(4x^3+5)(x^4-3x) is 28x^6-28x^3-15