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

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

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

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

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derivative of

y = \frac{8 x ^{2} + 4 x + 8}{x}

\frac{12 x ^{2} + 2 x - 4}{x x}

Solution steps

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

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

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

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

Simplify

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

= \frac{12 x ^{2} + 2 x - 4}{x x}

\int_{- 1}^{1} t^{3} (1 + t^{4})^{3} d t

\frac{d}{d x} (\frac{e ^{x}}{ln ( x )})

\int 1 - \frac{x}{2} d x

\int \frac{ln ( x )}{2 x ^{2}} d x

5 (- csc^{2} (x) + sec^{2} (x))

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