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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (1) = 0

= - \frac{1}{( x + 1 ) ^{2}} + 0

Simplify

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- x} sec (2 x)

\frac{d ^{2}}{d x ^{2}} (6 csc (x))

\frac{d}{d x} (x sinh (x^{2}))

\int \frac{1}{x ^{3} x ^{2} - 16} d x

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