What is the derivative of y= 1/(x+\frac{1){x+1}} ?

The derivative of y= 1/(x+\frac{1){x+1}} is -(x^2+2x)/((x(x+1)+1)^2)

What is the first derivative of y= 1/(x+\frac{1){x+1}} ?

The first derivative of y= 1/(x+\frac{1){x+1}} is -(x^2+2x)/((x(x+1)+1)^2)

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derivative of

y = \frac{1}{x + \frac{1}{x + 1}}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((3 x)^{- 4})

8 e^{x} + x

f (x) = \frac{x ^{8} + 2 x + 8}{x ^{2}}

y^{^{'}} = 25 - 10 t - 2 y

\int \frac{1}{64 e ^{- 2 x} + e ^{2 x}} d x

What is the derivative of y= 1/(x+\frac{1){x+1}} ?
The derivative of y= 1/(x+\frac{1){x+1}} is -(x^2+2x)/((x(x+1)+1)^2)
What is the first derivative of y= 1/(x+\frac{1){x+1}} ?
The first derivative of y= 1/(x+\frac{1){x+1}} is -(x^2+2x)/((x(x+1)+1)^2)