Question

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

Answer 1

Answer:

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.

Answer 2

Answer:

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