What is the derivative of (3x^2+9/(6x)) ?

The derivative of (3x^2+9/(6x)) is (x^2-3)/(2x^2)

What is the first derivative of (3x^2+9/(6x)) ?

The first derivative of (3x^2+9/(6x)) is (x^2-3)/(2x^2)

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\frac{d}{d x} (\frac{3 x ^{2} + 9}{6 x})

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

Solution steps

\frac{d}{d x} (\frac{3 x ^{2} + 9}{6 x})

Take the constant out:

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

= \frac{1}{6} \frac{d}{d x} (\frac{3 x ^{2} + 9}{x})

Apply the Quotient Rule:

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

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( 3 x ^{2} + 9 ) x - \frac{d x}{d x} ( 3 x ^{2} + 9 )}{x ^{2}}

\frac{d}{d x} (3 x^{2} + 9) = 6 x

\frac{d x}{d x} = 1

= \frac{1}{6} \cdot \frac{6 xx - 1 \cdot ( 3 x ^{2} + 9 )}{x ^{2}}

Simplify

\frac{1}{6} \cdot \frac{6 xx - 1 \cdot ( 3 x ^{2} + 9 )}{x ^{2}} : \frac{x ^{2} - 3}{2 x ^{2}}

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

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

e^{5 x}

\int_{0}^{1} \frac{1}{x ^{2}} d x

\int \frac{tan ^{2} ( x )}{sec ^{2} ( x )} d x

\int \frac{2 x + 1}{x - 2} d x

What is the derivative of (3x^2+9/(6x)) ?
The derivative of (3x^2+9/(6x)) is (x^2-3)/(2x^2)
What is the first derivative of (3x^2+9/(6x)) ?
The first derivative of (3x^2+9/(6x)) is (x^2-3)/(2x^2)