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

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

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

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

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derivative of

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

3 x^{2} + 2 x + 1

Solution steps

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

Apply the Product Rule:

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

f = x^{2} + 1, g = x + 1

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

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

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

= 2 x (x + 1) + 1 \cdot (x^{2} + 1)

Simplify

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

= 3 x^{2} + 2 x + 1

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

arctan (x^{2} + 1)

y = \frac{( x - 3 ) ^{2} - 4}{x ^{2} + 2}

\frac{y ^{2}}{x}

\frac{x ^{2} - 5}{1 - 3 x ^{2}}

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