What is the derivative of (3x^2+4x+8sqrt(x-1)) ?

The derivative of (3x^2+4x+8sqrt(x-1)) is (15x^2)/(2sqrt(x-1))

What is the first derivative of (3x^2+4x+8sqrt(x-1)) ?

The first derivative of (3x^2+4x+8sqrt(x-1)) is (15x^2)/(2sqrt(x-1))

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

\frac{15 x ^{2}}{2 x - 1}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 x^{2} + 4 x + 8, g = x - 1

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

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

\frac{d}{d x} (x - 1) = \frac{1}{2 x - 1}

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

Simplify

(6 x + 4) x - 1 + \frac{1}{2 x - 1} (3 x^{2} + 4 x + 8) : \frac{15 x ^{2}}{2 x - 1}

= \frac{15 x ^{2}}{2 x - 1}

n = 3 \sum \infty \frac{8}{n ^{2} - 1}

4 x^{3} + 6 x^{2}

\frac{\partial}{\partial x} (6 x + 3 y - 7)

\frac{d}{d y} (2 e^{y})

x y + y^{^{'}} = 78 x

What is the derivative of (3x^2+4x+8sqrt(x-1)) ?
The derivative of (3x^2+4x+8sqrt(x-1)) is (15x^2)/(2sqrt(x-1))
What is the first derivative of (3x^2+4x+8sqrt(x-1)) ?
The first derivative of (3x^2+4x+8sqrt(x-1)) is (15x^2)/(2sqrt(x-1))