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

The derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))

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

The first derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))

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derivative of

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

\frac{9 x ^{2} + 4 x - 1}{x x}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (6 x^{2} + 8 x + 2) = 12 x + 8

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

= \frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}} : \frac{9 x ^{2} + 4 x - 1}{x x}

= \frac{9 x ^{2} + 4 x - 1}{x x}

\int \frac{s ( s + 1 ) ^{2}}{s} d s

(sin (x) cos (x))^{^{'}}

\int_{1}^{2} 5 x d x

f (x) = 2^{x} 1

What is the derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) ?
The derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))
What is the first derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) ?
The first derivative of f(x)=(6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))