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

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

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

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

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derivative of

f (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}}

5 - x^{2}, sin (x), - 2.38468, 2.02521

7 e^{3 t} - 8 e^{- t} + 3

\int tan^{- 3} (x) sec^{4} (x) d x

\int_{- 10}^{x} t^{2} + 11 d t

f (x) = \frac{8}{x}, at x = 1

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