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

The derivative of sqrt(2x^2+8) is (2x)/(sqrt(2x^2+8))

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

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

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derivative of

2 x^{2} + 8

\frac{2 x}{2 x ^{2} + 8}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (2 x^{2} + 8) = 4 x

= \frac{1}{2 2 x ^{2} + 8} \cdot 4 x

Simplify

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

= \frac{2 x}{2 x ^{2} + 8}

f (x) = 6 e^{x} + 5 x - ln (x)

\int (x^{3} + 3) x^{2} d x

\frac{d}{d x} (y^{2} (6 - 2 x))

\int \frac{1}{x 10 x + 81} d x

n = 0 \sum \infty \frac{n ^{2}}{n ^{3}}

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