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

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

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

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

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

\frac{3}{x ^{4}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{3}{x ^{4}}

f (x) = 3 - 2 x + 4 x^{3}, at x = - 1

\int \frac{1}{e ^{\frac{x}{3}} + e ^{\frac{x}{6}}} d x

\frac{d}{d x} (7 e^{8 x})

f (x) = 25 - x^{2}

\frac{d}{d x} (\frac{50 x ^{4}}{2 ^{x}})

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