What is the derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) ?

The derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) is (12x^2+3x-2)/(xsqrt(x))

What is the first derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) ?

The first derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) is (12x^2+3x-2)/(xsqrt(x))

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derivative of

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

\frac{12 x ^{2} + 3 x - 2}{x x}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (8 x^{2} + 6 x + 4) = 16 x + 6

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

= \frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 4 )}{( x ) ^{2}} : \frac{12 x ^{2} + 3 x - 2}{x x}

= \frac{12 x ^{2} + 3 x - 2}{x x}

\frac{d y}{d t} = t 1 - y^{2}

\frac{\partial}{\partial x} (4 x^{3} - 4 yz)

\frac{\partial}{\partial x} (e^{y} - y e^{x})

(a - 1) + 1

What is the derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) ?
The derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) is (12x^2+3x-2)/(xsqrt(x))
What is the first derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) ?
The first derivative of f(x)=(8x^2+6x+4)/(sqrt(x)) is (12x^2+3x-2)/(xsqrt(x))