What is the derivative of sqrt(3x^2+4x+8) ?

The derivative of sqrt(3x^2+4x+8) is (3x+2)/(sqrt(3x^2+4x+8))

What is the first derivative of sqrt(3x^2+4x+8) ?

The first derivative of sqrt(3x^2+4x+8) is (3x+2)/(sqrt(3x^2+4x+8))

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derivative of

3 x^{2} + 4 x + 8

\frac{3 x + 2}{3 x ^{2} + 4 x + 8}

Solution steps

\frac{d}{d x} (3 x^{2} + 4 x + 8)

Apply the chain rule:

\frac{1}{2 3 x ^{2} + 4 x + 8} \frac{d}{d x} (3 x^{2} + 4 x + 8)

= \frac{1}{2 3 x ^{2} + 4 x + 8} \frac{d}{d x} (3 x^{2} + 4 x + 8)

\frac{d}{d x} (3 x^{2} + 4 x + 8) = 6 x + 4

= \frac{1}{2 3 x ^{2} + 4 x + 8} (6 x + 4)

Simplify

\frac{1}{2 3 x ^{2} + 4 x + 8} (6 x + 4) : \frac{3 x + 2}{3 x ^{2} + 4 x + 8}

= \frac{3 x + 2}{3 x ^{2} + 4 x + 8}

(x^{2} + 7)^{5}

x, x^{3} - x^{2} - x

2 sin (x), 4 cos (x), [0, 0.6 π]

\frac{d}{d x} ((x^{5} - x^{3} + 3)^{4})

3 y^{^{'}} + x y = 5 x

What is the derivative of sqrt(3x^2+4x+8) ?
The derivative of sqrt(3x^2+4x+8) is (3x+2)/(sqrt(3x^2+4x+8))
What is the first derivative of sqrt(3x^2+4x+8) ?
The first derivative of sqrt(3x^2+4x+8) is (3x+2)/(sqrt(3x^2+4x+8))