What is the derivative of (2x+8^2) ?

The derivative of (2x+8^2) is 4(2x+8)

What is the first derivative of (2x+8^2) ?

The first derivative of (2x+8^2) is 4(2x+8)

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\frac{d}{d x} ((2 x + 8)^{2})

4 (2 x + 8)

Solution steps

\frac{d}{d x} ((2 x + 8)^{2})

Apply the chain rule:

2 (2 x + 8) \frac{d}{d x} (2 x + 8)

= 2 (2 x + 8) \frac{d}{d x} (2 x + 8)

\frac{d}{d x} (2 x + 8) = 2

= 2 (2 x + 8) \cdot 2

Simplify

= 4 (2 x + 8)

\int 2 x^{3} d x

(\frac{2}{t})^{^{'}}

\frac{1}{5 x + 7}

y^{^{'}} = 2 y - y^{3}

\int \frac{1}{x ^{2} - 3} d x

What is the derivative of (2x+8^2) ?
The derivative of (2x+8^2) is 4(2x+8)
What is the first derivative of (2x+8^2) ?
The first derivative of (2x+8^2) is 4(2x+8)