What is the derivative of y=(x^2+8x+3)/(sqrt(x)) ?

The derivative of y=(x^2+8x+3)/(sqrt(x)) is (3x^2+8x-3)/(2xsqrt(x))

What is the first derivative of y=(x^2+8x+3)/(sqrt(x)) ?

The first derivative of y=(x^2+8x+3)/(sqrt(x)) is (3x^2+8x-3)/(2xsqrt(x))

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derivative of

y = \frac{x ^{2} + 8 x + 3}{x}

\frac{3 x ^{2} + 8 x - 3}{2 x x}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 8 x + 3}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 8 x + 3 ) x - \frac{d}{d x} ( x ) ( x ^{2} + 8 x + 3 )}{( x ) ^{2}}

\frac{d}{d x} (x^{2} + 8 x + 3) = 2 x + 8

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

= \frac{( 2 x + 8 ) x - \frac{1}{2 x} ( x ^{2} + 8 x + 3 )}{( x ) ^{2}}

Simplify

\frac{( 2 x + 8 ) x - \frac{1}{2 x} ( x ^{2} + 8 x + 3 )}{( x ) ^{2}} : \frac{3 x ^{2} + 8 x - 3}{2 x x}

= \frac{3 x ^{2} + 8 x - 3}{2 x x}

\frac{d}{d x} (\frac{g ( x ) ( x )}{a x ^{2} + x})

t an g e n t \frac{7}{x ^{2} + 3}

\frac{d}{d x} (x^{2} sin (x) + 2 x cos (x))

\frac{d y}{d x} = \frac{2 y}{2 x + 1}, y (0) = e

\int_{0}^{4} x e^{x^{2}} d x

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