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

The derivative of 6(2x^3+3)6x^2 is 36(10x^4+6x)

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

The first derivative of 6(2x^3+3)6x^2 is 36(10x^4+6x)

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derivative of

6 (2 x^{3} + 3) 6 x^{2}

36 (10 x^{4} + 6 x)

Solution steps

\frac{d}{d x} (6 (2 x^{3} + 3) \cdot 6 x^{2})

Simplify

6 (2 x^{3} + 3) \cdot 6 x^{2} : 36 x^{2} (2 x^{3} + 3)

= \frac{d}{d x} (36 x^{2} (2 x^{3} + 3))

Take the constant out:

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

= 36 \frac{d}{d x} (x^{2} (2 x^{3} + 3))

Apply the Product Rule:

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

f = x^{2}, g = 2 x^{3} + 3

= 36 (\frac{d}{d x} (x^{2}) (2 x^{3} + 3) + \frac{d}{d x} (2 x^{3} + 3) x^{2})

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

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

= 36 (2 x (2 x^{3} + 3) + 6 x^{2} x^{2})

Simplify

36 (2 x (2 x^{3} + 3) + 6 x^{2} x^{2}) : 36 (10 x^{4} + 6 x)

= 36 (10 x^{4} + 6 x)

\int \frac{2 - 4 x ^{2}}{x - x ^{3}} d x

y = F - x^{2} - x, at x = - 2

f (x) = 10 t

y^{^{'}} = x + 4 y

n = 0 \sum \infty 2 n

What is the derivative of 6(2x^3+3)6x^2 ?
The derivative of 6(2x^3+3)6x^2 is 36(10x^4+6x)
What is the first derivative of 6(2x^3+3)6x^2 ?
The first derivative of 6(2x^3+3)6x^2 is 36(10x^4+6x)