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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derivative of

\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 \frac{1}{2 x ^{2} + 4 x + 5} d x

\frac{d}{d x} (\frac{x ^{2} + 6}{sin ^{4} ( x )})

f (x) = - 9 e^{x^{- 9}}

\frac{d}{d x} (- \frac{2}{1 - 2 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