What is the derivative of f(x)=((x^3+4x-1))/(x^2) ?

The derivative of f(x)=((x^3+4x-1))/(x^2) is (x^3-4x+2)/(x^3)

What is the first derivative of f(x)=((x^3+4x-1))/(x^2) ?

The first derivative of f(x)=((x^3+4x-1))/(x^2) is (x^3-4x+2)/(x^3)

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derivative of

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

\frac{x ^{3} - 4 x + 2}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 4 x - 1}{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} + 4 x - 1 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{3} + 4 x - 1 )}{( x ^{2} ) ^{2}}

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

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

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

Simplify

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

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

d er i v a t i v e \frac{( 1 + 2 x + x ^{2} )}{e ^{x}}

d er i v a t i v e \frac{x ^{2}}{2 + 3 x}

d er i v a t i v e 4 x ln (2 x + 1)

\frac{d}{d q} (2600 - 13 q)

d er i v a t i v e cos (x) + 1

What is the derivative of f(x)=((x^3+4x-1))/(x^2) ?
The derivative of f(x)=((x^3+4x-1))/(x^2) is (x^3-4x+2)/(x^3)
What is the first derivative of f(x)=((x^3+4x-1))/(x^2) ?
The first derivative of f(x)=((x^3+4x-1))/(x^2) is (x^3-4x+2)/(x^3)