Find the dervite of y=X^2 + x + 1

y' (x) = 1 + 2 x

Step-by-step explanation:

Find the derivative of the following via implicit differentiation:

d/dx (y) = d/dx (1 + x + x^2)

Using the chain rule, d/dx (y) = (dy (u)) / (du) (du) / (dx), where u = x and d / (du) (y (u)) = y' (u):

d/dx (x) y' (x) = d/dx (1 + x + x^2)

The derivative of x is 1:

1 y' (x) = d/dx (1 + x + x^2)

Differentiate the sum term by term:

y' (x) = d/dx (1) + d/dx (x) + d/dx (x^2)

The derivative of 1 is zero:

y' (x) = d/dx (x) + d/dx (x^2) + 0

Simplify the expression:

y' (x) = d/dx (x) + d/dx (x^2)

The derivative of x is 1:

y' (x) = d/dx (x^2) + 1

Use the power rule, d/dx (x^n) = n x^ (n - 1), where n = 2.

d/dx (x^2) = 2 x:

Answer: y' (x) = 1 + 2 x

2x + 1.

Step-by-step explanation:

y=x^2 + x + 1

Using the algebraic derivative rule, if y = ax^n then y' = anx^ (n-1):

The derivative is 2x^ (2 - 1) + 1x^ (1-1)

= 2x^1 + x^0

= 2x + 1.