Find f'(x) if f(x) = (ln(x^2+x+1)+2x+3)\
(x^2+1) .
To find the derivative of f(x), we can use the product rule. The product rule states that if we have a function f(x) = g(x) * h(x), then the derivative of f(x) is f'(x) = g'(x) * h(x) + g(x) * h'(x).
In this case, we have f(x) = (ln(x^2+x+1)+2x+3) * (x^2+1). Let's find the derivative step by step:
Step 1: Find the derivative of the first factor, g(x) = ln(x^2+x+1) + 2x + 3.
To find the derivative of ln(x^2+x+1), we need to apply the chain rule. The derivative of ln(u) is (1/u) * u', where u is the expression inside the natural logarithm.
So, the derivative of ln(x^2+x+1) is [(1/(x^2+x+1)) * (2x+1)]. Since we have an additional term 2x+3 in g(x), its derivative is simply 2.
Therefore, g'(x) = [(1/(x^2+x+1)) * (2x+1)] + 2.
Step 2: Find the derivative of the second factor, h(x) = x^2 + 1.
The derivative of x^2 + 1 is simply 2x, as the derivative of x^2 is 2x and the derivative of a constant (1) is zero.
Therefore, h'(x) = 2x.
Step 3: Apply the product rule.
Using the product rule, f'(x) = g'(x) * h(x) + g(x) * h'(x).
Plugging in the derivatives we found in steps 1 and 2, we have:
f'(x) = [(1/(x^2+x+1)) * (2x+1) + 2] * (x^2 + 1) + (ln(x^2+x+1) + 2x + 3) * 2x.
Simplifying this expression further may be difficult without specific values of x, but this is the derivative of f(x) using the product rule.