What is the derivative of (x^3/(sqrt(x)+x)) ?

The derivative of (x^3/(sqrt(x)+x)) is (4x^3+6x^2sqrt(x)-x^{5/2})/(2(sqrt(x)+x)^2)

What is the first derivative of (x^3/(sqrt(x)+x)) ?

The first derivative of (x^3/(sqrt(x)+x)) is (4x^3+6x^2sqrt(x)-x^{5/2})/(2(sqrt(x)+x)^2)

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

\frac{4 x ^{3} + 6 x ^{2} x - x ^{\frac{5}{2}}}{2 ( x + x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x + x) = \frac{1}{2 x} + 1

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

Simplify

\frac{3 x ^{2} ( x + x ) - ( \frac{1}{2 x} + 1 ) x ^{3}}{( x + x ) ^{2}} : \frac{4 x ^{3} + 6 x ^{2} x - x ^{\frac{5}{2}}}{2 ( x + x ) ^{2}}

= \frac{4 x ^{3} + 6 x ^{2} x - x ^{\frac{5}{2}}}{2 ( x + x ) ^{2}}

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

x \to 0 + lim (x^{x^{4}})

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

\frac{\partial}{\partial x} (1 + 2 x y)

\frac{s}{( s + 2 ) ^{2} + 1}

What is the derivative of (x^3/(sqrt(x)+x)) ?
The derivative of (x^3/(sqrt(x)+x)) is (4x^3+6x^2sqrt(x)-x^{5/2})/(2(sqrt(x)+x)^2)
What is the first derivative of (x^3/(sqrt(x)+x)) ?
The first derivative of (x^3/(sqrt(x)+x)) is (4x^3+6x^2sqrt(x)-x^{5/2})/(2(sqrt(x)+x)^2)