What is the derivative of (x^2+x+1/(2sqrt(x))) ?

The derivative of (x^2+x+1/(2sqrt(x))) is (3x^2+x-1)/(4xsqrt(x))

What is the first derivative of (x^2+x+1/(2sqrt(x))) ?

The first derivative of (x^2+x+1/(2sqrt(x))) is (3x^2+x-1)/(4xsqrt(x))

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

\frac{3 x ^{2} + x - 1}{4 x x}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( x ^{2} + x + 1 ) x - \frac{d}{d x} ( x ) ( x ^{2} + x + 1 )}{( x ) ^{2}}

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

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

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

Simplify

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

= \frac{3 x ^{2} + x - 1}{4 x x}

\frac{d y}{d x} = 2 x + y, y (0) = 4

\frac{d}{d x} (6 e^{- 8 x})

f (x) = 2 x + 1, (4, 3)

(x^{3} - 4 x) (2 x^{2} + 5 x)

\int e^{t (a - s)} d t

What is the derivative of (x^2+x+1/(2sqrt(x))) ?
The derivative of (x^2+x+1/(2sqrt(x))) is (3x^2+x-1)/(4xsqrt(x))
What is the first derivative of (x^2+x+1/(2sqrt(x))) ?
The first derivative of (x^2+x+1/(2sqrt(x))) is (3x^2+x-1)/(4xsqrt(x))