What is the derivative of (8x+1^2) ?

The derivative of (8x+1^2) is 16(8x+1)

What is the first derivative of (8x+1^2) ?

The first derivative of (8x+1^2) is 16(8x+1)

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\frac{d}{d x} ((8 x + 1)^{2})

16 (8 x + 1)

Solution steps

\frac{d}{d x} ((8 x + 1)^{2})

Apply the chain rule:

2 (8 x + 1) \frac{d}{d x} (8 x + 1)

= 2 (8 x + 1) \frac{d}{d x} (8 x + 1)

\frac{d}{d x} (8 x + 1) = 8

= 2 (8 x + 1) \cdot 8

Simplify

= 16 (8 x + 1)

\int \frac{17}{( x + 1 ) ( x ^{2} + 16 )} d x

\int sec^{3} (θ) d θ

\frac{\partial}{\partial x} (x^{3} - 2 x^{2} + y^{3} - 2 y)

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

\int \frac{x ^{2} + 5 x ^{3} - 7}{x ^{2}} d x

What is the derivative of (8x+1^2) ?
The derivative of (8x+1^2) is 16(8x+1)
What is the first derivative of (8x+1^2) ?
The first derivative of (8x+1^2) is 16(8x+1)