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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int \frac{1 + sin ( x )}{cos ( x )} d x

\int_{0}^{9} x^{2} + 81 d x

\int \frac{( x - 2 ) ^{4}}{4 x - x ^{2}} d x

y = ∣ 10 x ∣, y = x^{2} - 11

\frac{d x}{d t} = 2 \frac{x}{t} - 0.194 t

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