What is the derivative of y=(sqrt(x)+x^3)/(x^2) ?

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

What is the first derivative of y=(sqrt(x)+x^3)/(x^2) ?

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = 7 - 2 x

\frac{\partial}{\partial x} (2 x^{3} + 4 y - 2 z^{2})

\int 2 \cdot x \cdot ln (4 x) d x

\int y x d x

\int_{0}^{π} sin^{2} (t co) s^{4} t d t

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