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
