What is the derivative of y=(x+1/x)(x-1/x+1) ?

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

What is the first derivative of y=(x+1/x)(x-1/x+1) ?

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

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derivative of

y = (x + \frac{1}{x}) (x - \frac{1}{x} + 1)

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

Solution steps

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

Apply the Product Rule:

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

f = x + \frac{1}{x}, g = x - \frac{1}{x} + 1

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (x^{2} + \frac{x}{y})

\frac{d}{d x} (2 ln (x))

\int_{0}^{1} 9 e^{- 4 x} d x

\frac{d}{d x} ((e^{x} + 1)^{7 x})

\frac{\partial}{\partial x} (- x + y + 2)

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