What is the derivative of (2x^2-5x+4/(sqrt(x))) ?

The derivative of (2x^2-5x+4/(sqrt(x))) is (6x^2-5x-4)/(2xsqrt(x))

What is the first derivative of (2x^2-5x+4/(sqrt(x))) ?

The first derivative of (2x^2-5x+4/(sqrt(x))) is (6x^2-5x-4)/(2xsqrt(x))

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\frac{d}{d x} (\frac{2 x ^{2} - 5 x + 4}{x})

\frac{6 x ^{2} - 5 x - 4}{2 x x}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} - 5 x + 4}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 x ^{2} - 5 x + 4 ) x - \frac{d}{d x} ( x ) ( 2 x ^{2} - 5 x + 4 )}{( x ) ^{2}}

\frac{d}{d x} (2 x^{2} - 5 x + 4) = 4 x - 5

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

= \frac{( 4 x - 5 ) x - \frac{1}{2 x} ( 2 x ^{2} - 5 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 4 x - 5 ) x - \frac{1}{2 x} ( 2 x ^{2} - 5 x + 4 )}{( x ) ^{2}} : \frac{6 x ^{2} - 5 x - 4}{2 x x}

= \frac{6 x ^{2} - 5 x - 4}{2 x x}

y = x + 4, y = \frac{x}{2} + 2

\int 8 tan^{4} (x se) c^{6} x d x

e^{\frac{1}{2} x}

4 - 4 cos (7 x), 4 cos (7 x), 0 \geq x \geq \frac{π}{7}

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

What is the derivative of (2x^2-5x+4/(sqrt(x))) ?
The derivative of (2x^2-5x+4/(sqrt(x))) is (6x^2-5x-4)/(2xsqrt(x))
What is the first derivative of (2x^2-5x+4/(sqrt(x))) ?
The first derivative of (2x^2-5x+4/(sqrt(x))) is (6x^2-5x-4)/(2xsqrt(x))