What is the derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) ?

The derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) is (x^3-4x+10)/(x^3)

What is the first derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) ?

The first derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) is (x^3-4x+10)/(x^3)

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derivative of

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

\frac{x ^{3} - 4 x + 10}{x ^{3}}

Solution steps

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

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

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

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

Simplify

\frac{( 3 x ^{2} - 6 x + 4 ) x ^{2} - 2 x ( x ^{3} - 3 x ^{2} + 4 x - 5 )}{( x ^{2} ) ^{2}} : \frac{x ^{3} - 4 x + 10}{x ^{3}}

= \frac{x ^{3} - 4 x + 10}{x ^{3}}

f (x) = e^{- x^{2}} + x

y^{^{'}} + 4 y = x^{2} e^{- 4 x}

3 t + 1

(sin (4 x))^{^{'}}

\int_{0}^{b} \frac{1}{9 - x ^{2}} d x

What is the derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) ?
The derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) is (x^3-4x+10)/(x^3)
What is the first derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) ?
The first derivative of F(x)=(x^3-3x^2+4x-5)/(x^2) is (x^3-4x+10)/(x^3)