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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{13}{( x - 7 ) ( x + 6 )} d x

\int e^{- \frac{x}{y}} d x

\frac{d}{d x} (a x a + x)

f (x) = tan (x^{2})

x \to - 4 lim (\frac{x ^{2}}{4 - x})

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