What is the derivative of (4x^2+6x+8/(sqrt(x))) ?

The derivative of (4x^2+6x+8/(sqrt(x))) is (6x^2+3x-4)/(xsqrt(x))

What is the first derivative of (4x^2+6x+8/(sqrt(x))) ?

The first derivative of (4x^2+6x+8/(sqrt(x))) is (6x^2+3x-4)/(xsqrt(x))

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\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 8}{x})

\frac{6 x ^{2} + 3 x - 4}{x x}

Solution steps

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 8}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{2} + 6 x + 8 ) x - \frac{d}{d x} ( x ) ( 4 x ^{2} + 6 x + 8 )}{( x ) ^{2}}

\frac{d}{d x} (4 x^{2} + 6 x + 8) = 8 x + 6

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

= \frac{( 8 x + 6 ) x - \frac{1}{2 x} ( 4 x ^{2} + 6 x + 8 )}{( x ) ^{2}}

Simplify

\frac{( 8 x + 6 ) x - \frac{1}{2 x} ( 4 x ^{2} + 6 x + 8 )}{( x ) ^{2}} : \frac{6 x ^{2} + 3 x - 4}{x x}

= \frac{6 x ^{2} + 3 x - 4}{x x}

\int \frac{y}{1 + x ^{2}} d y

f (x) = 4 + 4 x^{2} - 2 x^{3}, at x = a

t \to 0 lim (sin (6 t))

d y - (2 y - e^{2 x}) d x = 0

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

What is the derivative of (4x^2+6x+8/(sqrt(x))) ?
The derivative of (4x^2+6x+8/(sqrt(x))) is (6x^2+3x-4)/(xsqrt(x))
What is the first derivative of (4x^2+6x+8/(sqrt(x))) ?
The first derivative of (4x^2+6x+8/(sqrt(x))) is (6x^2+3x-4)/(xsqrt(x))