What is the derivative of (x^3-\sqrt[4]{x^3}/(2x)) ?

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

What is the first derivative of (x^3-\sqrt[4]{x^3}/(2x)) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

\frac{d}{d x} (x^{3} - 4 x^{3}) = \frac{- 3 + 12 x ^{\frac{9}{4}}}{4 x ^{\frac{1}{4}}}

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

= \frac{1}{2} \cdot \frac{\frac{- 3 + 12 x ^{\frac{9}{4}}}{4 x ^{\frac{1}{4}}} x - 1 \cdot ( x ^{3} - 4 x ^{3} )}{x ^{2}}

Simplify

\frac{1}{2} \cdot \frac{\frac{- 3 + 12 x ^{\frac{9}{4}}}{4 x ^{\frac{1}{4}}} x - 1 \cdot ( x ^{3} - 4 x ^{3} )}{x ^{2}} : \frac{8 x ^{3} - 3 x ^{\frac{3}{4}} + 4 4 x ^{3}}{8 x ^{2}}

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

n = 1 \sum \infty 9^{n} x^{n} n!

x \to 100 lim (7)

\frac{\partial}{\partial x} (e^{a x + b y})

\frac{d}{d x} (e^{x} + e^{xx})

\frac{d}{d x} (0.5 x^{3} + 1.5)

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