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

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

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

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

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derivative of

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

1

Solution steps

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

Simplify

\frac{x ^{2} + 2 x + 1}{x + 1} : x + 1

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

Apply the Sum/Difference Rule:

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

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

\frac{d x}{d x} = 1

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

= 1 + 0

Simplify

= 1

y = x^{4} + 12 x

\frac{d}{d a} (tan^{2} (a) + sin (a) csc (a))

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

(x^{2} - x + 1) e^{3} x

y = 6 x^{2} - 3 x^{3}

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