Now, what makes the numerator = 0?.
0 is obviously one solution.
Hello, I'm having difficulty with the following question:
Solve the equation f'(x)=0.
f(x)= (x^2 + 1)^-1 ln (x^2 + 1)
I've worked out the derivative to be ln(x+1) divided by (x^2 + 1)^2. However, I'm unsure if this is correct, and if it is, I don't know how to solve for the x values once I set the derivative equal to zero- the natural log in the numerator throws me off. Thank you for any help!