What is the derivative of (4x^{3/2}-x^3+8/x) ?

The derivative of (4x^{3/2}-x^3+8/x) is (-2x^3+2x^{3/2}-8)/(x^2)

What is the first derivative of (4x^{3/2}-x^3+8/x) ?

The first derivative of (4x^{3/2}-x^3+8/x) is (-2x^3+2x^{3/2}-8)/(x^2)

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Popular >

\frac{d}{d x} (\frac{4 x ^{\frac{3}{2}} - x ^{3} + 8}{x})

\frac{- 2 x ^{3} + 2 x ^{\frac{3}{2}} - 8}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{\frac{3}{2}} - x ^{3} + 8}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{\frac{3}{2}} - x ^{3} + 8 ) x - \frac{d x}{d x} ( 4 x ^{\frac{3}{2}} - x ^{3} + 8 )}{x ^{2}}

\frac{d}{d x} (4 x^{\frac{3}{2}} - x^{3} + 8) = - 3 x^{2} + 6 x^{\frac{1}{2}}

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

= \frac{( - 3 x ^{2} + 6 x ^{\frac{1}{2}} ) x - 1 \cdot ( 4 x ^{\frac{3}{2}} - x ^{3} + 8 )}{x ^{2}}

Simplify

\frac{( - 3 x ^{2} + 6 x ^{\frac{1}{2}} ) x - 1 \cdot ( 4 x ^{\frac{3}{2}} - x ^{3} + 8 )}{x ^{2}} : \frac{- 2 x ^{3} + 2 x ^{\frac{3}{2}} - 8}{x ^{2}}

= \frac{- 2 x ^{3} + 2 x ^{\frac{3}{2}} - 8}{x ^{2}}

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

\frac{d e}{d t}

\frac{2 - 5 s}{s ^{2} + 9}

\int \frac{1}{2} x^{2} d x

\frac{d}{d x} (sin (x^{2} + 2 x y))

What is the derivative of (4x^{3/2}-x^3+8/x) ?
The derivative of (4x^{3/2}-x^3+8/x) is (-2x^3+2x^{3/2}-8)/(x^2)
What is the first derivative of (4x^{3/2}-x^3+8/x) ?
The first derivative of (4x^{3/2}-x^3+8/x) is (-2x^3+2x^{3/2}-8)/(x^2)