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

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

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

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

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derivative of

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

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

Solution steps

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

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

Simplify

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

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

\frac{\partial}{\partial x} (x + 4 y)

\int \frac{- 2 x - 4}{x ^{3} + x ^{2} + x} d x

\frac{\partial}{\partial y} (tan (\frac{x}{y}))

x \to 0 lim (x ln (e^{x} - 1))

\int_{1}^{4} x e^{3 x^{2}} d x

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