What is the derivative of (5x^3-2x^2(6x^3)) ?

The derivative of (5x^3-2x^2(6x^3)) is 6(30x^5-10x^4)

What is the first derivative of (5x^3-2x^2(6x^3)) ?

The first derivative of (5x^3-2x^2(6x^3)) is 6(30x^5-10x^4)

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Popular >

\frac{d}{d x} ((5 x^{3} - 2 x^{2}) (6 x^{3}))

6 (30 x^{5} - 10 x^{4})

Solution steps

\frac{d}{d x} ((5 x^{3} - 2 x^{2}) (6 x^{3}))

Take the constant out:

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

= 6 \frac{d}{d x} ((5 x^{3} - 2 x^{2}) x^{3})

Apply the Product Rule:

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

f = 5 x^{3} - 2 x^{2}, g = x^{3}

= 6 (\frac{d}{d x} (5 x^{3} - 2 x^{2}) x^{3} + \frac{d}{d x} (x^{3}) (5 x^{3} - 2 x^{2}))

\frac{d}{d x} (5 x^{3} - 2 x^{2}) = 15 x^{2} - 4 x

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

= 6 ((15 x^{2} - 4 x) x^{3} + 3 x^{2} (5 x^{3} - 2 x^{2}))

Expand

(15 x^{2} - 4 x) x^{3} + 3 x^{2} (5 x^{3} - 2 x^{2}) : 30 x^{5} - 10 x^{4}

= 6 (30 x^{5} - 10 x^{4})

\frac{d}{d x} (1 - \frac{x ^{2}}{16})

y^{2} d x + x (x - y) d y = 0

\int_{- 1}^{1} 2 x d x

2 t^{2} - 3 t + 3

x \to \infty lim (5 x^{4})

What is the derivative of (5x^3-2x^2(6x^3)) ?
The derivative of (5x^3-2x^2(6x^3)) is 6(30x^5-10x^4)
What is the first derivative of (5x^3-2x^2(6x^3)) ?
The first derivative of (5x^3-2x^2(6x^3)) is 6(30x^5-10x^4)