What is the derivative of (3x^2-3/(2x)) ?

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

What is the first derivative of (3x^2-3/(2x)) ?

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

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Popular >

\frac{d}{d x} (\frac{3 x ^{2} - 3}{2 x})

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

Solution steps

\frac{d}{d x} (\frac{3 x ^{2} - 3}{2 x})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{3 x ^{2} - 3}{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} ( 3 x ^{2} - 3 ) x - \frac{d x}{d x} ( 3 x ^{2} - 3 )}{x ^{2}}

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

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

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

Simplify

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

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

\int 2 x^{2} e^{x} d x

y^{^{'}} = 1 + 9 x - 5 y, y (1) = 2

\frac{\partial}{\partial y} (ln (x^{2} + y^{2}))

\int \frac{9 x ^{2} + 5 x + 9}{( x ^{2} + 1 ) ^{2}} d x

ln (4 + x^{2})

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