What is the derivative of (a+bx+cx^2/(sqrt(x))) ?

The derivative of (a+bx+cx^2/(sqrt(x))) is (3cx^2+bx-a)/(2xsqrt(x))

What is the first derivative of (a+bx+cx^2/(sqrt(x))) ?

The first derivative of (a+bx+cx^2/(sqrt(x))) is (3cx^2+bx-a)/(2xsqrt(x))

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

\frac{3 c x ^{2} + b x - a}{2 x x}

Solution steps

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

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( a + b x + c x ^{2} ) x - \frac{d}{d x} ( x ) ( a + b x + c x ^{2} )}{( x ) ^{2}}

\frac{d}{d x} (a + b x + c x^{2}) = 2 c x + b

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

= \frac{( 2 c x + b ) x - \frac{1}{2 x} ( a + b x + c x ^{2} )}{( x ) ^{2}}

Simplify

\frac{( 2 c x + b ) x - \frac{1}{2 x} ( a + b x + c x ^{2} )}{( x ) ^{2}} : \frac{3 c x ^{2} + b x - a}{2 x x}

= \frac{3 c x ^{2} + b x - a}{2 x x}

\frac{d}{d x} (4 - cos (x))

x \to 2 + lim (4 - x^{2})

\frac{d}{d x} (e^{- x^{2}} - x)

(6.3) (9.6)

y^{^{''}} - 4 y^{^{'}} + 5 y = 0

What is the derivative of (a+bx+cx^2/(sqrt(x))) ?
The derivative of (a+bx+cx^2/(sqrt(x))) is (3cx^2+bx-a)/(2xsqrt(x))
What is the first derivative of (a+bx+cx^2/(sqrt(x))) ?
The first derivative of (a+bx+cx^2/(sqrt(x))) is (3cx^2+bx-a)/(2xsqrt(x))