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

The derivative of sqrt(8x+1) is 4/(sqrt(8x+1))

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

The first derivative of sqrt(8x+1) is 4/(sqrt(8x+1))

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

\frac{4}{8 x + 1}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{2 8 x + 1} \cdot 8

Simplify

\frac{1}{2 8 x + 1} \cdot 8 : \frac{4}{8 x + 1}

= \frac{4}{8 x + 1}

\int_{0}^{2} (x + 2)^{2} d x

f (x) = 6 x

\frac{d}{d x} (sinh (ln (x)))

n \to \infty lim (a n^{2})

y = (x^{2} + 8 x - 4) (2 x^{- 2} + x^{- 4})

What is the derivative of sqrt(8x+1) ?
The derivative of sqrt(8x+1) is 4/(sqrt(8x+1))
What is the first derivative of sqrt(8x+1) ?
The first derivative of sqrt(8x+1) is 4/(sqrt(8x+1))