What is the derivative of (x^6-1/(x-1)) ?

The derivative of (x^6-1/(x-1)) is 5x^4+4x^3+3x^2+2x+1

What is the first derivative of (x^6-1/(x-1)) ?

The first derivative of (x^6-1/(x-1)) is 5x^4+4x^3+3x^2+2x+1

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\frac{d}{d x} (\frac{x ^{6} - 1}{x - 1})

5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1

Solution steps

\frac{d}{d x} (\frac{x ^{6} - 1}{x - 1})

Simplify

\frac{x ^{6} - 1}{x - 1} : (x + 1) (x^{4} + x^{2} + 1)

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x + 1, g = x^{4} + x^{2} + 1

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

\frac{d}{d x} (x + 1) = 1

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

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

Simplify

1 \cdot (x^{4} + x^{2} + 1) + (4 x^{3} + 2 x) (x + 1) : 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1

= 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1

\int_{0}^{1} 3 x^{4} e^{- x} d x

y = - 5 x^{2}, (1, - 5)

20 ln (3 x + 1) - \frac{x}{3}

y = 8 x - x^{2}

\int \frac{1}{2 e ^{x}} d x

What is the derivative of (x^6-1/(x-1)) ?
The derivative of (x^6-1/(x-1)) is 5x^4+4x^3+3x^2+2x+1
What is the first derivative of (x^6-1/(x-1)) ?
The first derivative of (x^6-1/(x-1)) is 5x^4+4x^3+3x^2+2x+1