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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{1}{( s - 5 ) ^{4}}

y^{^{'}} + tan (2 \cdot x) y = 0

\int 2 x (x^{2} - 2) d x

f (x) = (x + 2) 2 x - 3, at x = 2

n = 1 \sum \infty (\frac{π}{5})^{n}

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