What is the derivative of (2x^{3/2}-x^2+4/x) ?

The derivative of (2x^{3/2}-x^2+4/x) is (-x^2+x^{3/2}-4)/(x^2)

What is the first derivative of (2x^{3/2}-x^2+4/x) ?

The first derivative of (2x^{3/2}-x^2+4/x) is (-x^2+x^{3/2}-4)/(x^2)

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\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}} - x ^{2} + 4}{x})

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

Solution steps

\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}} - x ^{2} + 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 ^{\frac{3}{2}} - x ^{2} + 4 ) x - \frac{d x}{d x} ( 2 x ^{\frac{3}{2}} - x ^{2} + 4 )}{x ^{2}}

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{1}{x ^{7}} + \frac{2}{x ^{3}})

\int_{- 1}^{1} 2 - 2 x^{2} d x

y^{^{''}} + 2 y^{^{'}} + y = x^{2} e^{- x}

tan (t)

f (x) = 36 x (9 - x^{2})^{- 2}

What is the derivative of (2x^{3/2}-x^2+4/x) ?
The derivative of (2x^{3/2}-x^2+4/x) is (-x^2+x^{3/2}-4)/(x^2)
What is the first derivative of (2x^{3/2}-x^2+4/x) ?
The first derivative of (2x^{3/2}-x^2+4/x) is (-x^2+x^{3/2}-4)/(x^2)