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

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

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

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

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derivative of

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

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

Solution steps

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

\frac{d}{d x} (4 x^{3} - 2 x + 1) = 12 x^{2} - 2

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

= \frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}}

Simplify

\frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}} : \frac{2 ( 2 x ^{3} + x - 1 )}{x ^{3}}

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

x \to + 6 lim (3)

\int x (1 - x)^{2} d x

x \to - 2 lim (x^{2} + 5 x + 4)

\int \frac{x + 2}{x + 1} d x

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

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