How would you find f'(1) if the function is f(x)= lnx^3 +x^2?

Answer 1

Let f(x) = ln(x³) + x²

Then the derivative of f(x), is given by

f'(x) = 3x²·(1/x³) + 2x

f'(x) = 3x^-1+ 2x

Now you can just plug in 1 to get f'(1)

f'(1) = 3·1 + 2·1 = 5

Answer 2

f(x) = ln(x^3) + x^2

the derivative is given by f'(x) = 1/x^3 * 3x^2 + 2x which we get by applying the chain rule

now we evaluate the derivative, which is itself a function, at the point x = 1

this gives us f'(1) = 1/(1)^3 * 3(1)^2 + 2(1) = 1/1*3 + 2 = 3 + 2 = 5 which is your answer

I hope that helps