What is the derivative of (3x^2-x+1/(x^7)) ?

The derivative of (3x^2-x+1/(x^7)) is (-15x^2+6x-7)/(x^8)

What is the first derivative of (3x^2-x+1/(x^7)) ?

The first derivative of (3x^2-x+1/(x^7)) is (-15x^2+6x-7)/(x^8)

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\frac{d}{d x} (\frac{3 x ^{2} - x + 1}{x ^{7}})

\frac{- 15 x ^{2} + 6 x - 7}{x ^{8}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{2} - x + 1}{x ^{7}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x ^{2} - x + 1 ) x ^{7} - \frac{d}{d x} ( x ^{7} ) ( 3 x ^{2} - x + 1 )}{( x ^{7} ) ^{2}}

\frac{d}{d x} (3 x^{2} - x + 1) = 6 x - 1

\frac{d}{d x} (x^{7}) = 7 x^{6}

= \frac{( 6 x - 1 ) x ^{7} - 7 x ^{6} ( 3 x ^{2} - x + 1 )}{( x ^{7} ) ^{2}}

Simplify

\frac{( 6 x - 1 ) x ^{7} - 7 x ^{6} ( 3 x ^{2} - x + 1 )}{( x ^{7} ) ^{2}} : \frac{- 15 x ^{2} + 6 x - 7}{x ^{8}}

= \frac{- 15 x ^{2} + 6 x - 7}{x ^{8}}

\int \frac{1}{x 36 x ^{2} + 1} d x

y^{^{'}} = - y^{\frac{1}{3}}

\frac{d}{d x} (2 sin (x) - x)

(- \frac{5}{2}, 1), (- \frac{1}{2}, 4)

\int \frac{5 x}{2} d x

What is the derivative of (3x^2-x+1/(x^7)) ?
The derivative of (3x^2-x+1/(x^7)) is (-15x^2+6x-7)/(x^8)
What is the first derivative of (3x^2-x+1/(x^7)) ?
The first derivative of (3x^2-x+1/(x^7)) is (-15x^2+6x-7)/(x^8)