What is the derivative of y=(3x^3+4)^2 ?

The derivative of y=(3x^3+4)^2 is 18x^2(3x^3+4)

What is the first derivative of y=(3x^3+4)^2 ?

The first derivative of y=(3x^3+4)^2 is 18x^2(3x^3+4)

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derivative of

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

18 x^{2} (3 x^{3} + 4)

Solution steps

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

Apply the chain rule:

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

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

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

= 2 (3 x^{3} + 4) \cdot 9 x^{2}

Simplify

= 18 x^{2} (3 x^{3} + 4)

\int 2^{x + 1} d x

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What is the derivative of y=(3x^3+4)^2 ?
The derivative of y=(3x^3+4)^2 is 18x^2(3x^3+4)
What is the first derivative of y=(3x^3+4)^2 ?
The first derivative of y=(3x^3+4)^2 is 18x^2(3x^3+4)