What is the derivative of (6x^2+8x+8x)/(sqrt(x)) ?

The derivative of (6x^2+8x+8x)/(sqrt(x)) is (9x+8)/(sqrt(x))

What is the first derivative of (6x^2+8x+8x)/(sqrt(x)) ?

The first derivative of (6x^2+8x+8x)/(sqrt(x)) is (9x+8)/(sqrt(x))

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derivative of

\frac{6 x ^{2} + 8 x + 8 x}{x}

\frac{9 x + 8}{x}

Solution steps

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

Simplify

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

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

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

f = x, g = 3 x + 8

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

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

\frac{d}{d x} (3 x + 8) = 3

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

Simplify

2 (\frac{1}{2 x} (3 x + 8) + 3 x) : \frac{9 x + 8}{x}

= \frac{9 x + 8}{x}

y^{^{'}} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

\int e^{3 x} sin (3 x) d x

y^{^{'}} = \frac{x}{y}, y (0) = 3

\frac{d}{d x} (\frac{18 x + 4 x ^{2}}{( 9 + 4 x ) ^{2}})

\int sin (t^{2}) d t

What is the derivative of (6x^2+8x+8x)/(sqrt(x)) ?
The derivative of (6x^2+8x+8x)/(sqrt(x)) is (9x+8)/(sqrt(x))
What is the first derivative of (6x^2+8x+8x)/(sqrt(x)) ?
The first derivative of (6x^2+8x+8x)/(sqrt(x)) is (9x+8)/(sqrt(x))