What is f′(x) given that f(x)=x2lnx−x ? (1 point) Responses f′(x)=2xlnx f ′ ( x ) = 2 x ln x f′(x)=xlnx+x−1 f ' ( x ) = x ln x + x − 1 f′(x)=2xlnx−x−1 f ′ ( x ) = 2 x ln x − x − 1 f′(x)=2xlnx+x−1 f ′ ( x ) = 2 x ln x + x − 1 f′(x)=2xlnx+2x−1 f ′ ( x ) = 2 x ln x + 2 x − 1 Question 2 Part B What is f′′(x) ? (1 point) Responses f′′(x)=2lnx+1 f '' ( x ) = 2 ln x + 1 f′′(x)=2lnx+2 f '' ( x ) = 2 ln x + 2 f′′(x)=2lnx+3 f '' ( x ) = 2 ln x + 3 f′′(x)=2lnx+2x+1 f '' ( x ) = 2 ln x + 2 x + 1 f′′(x)=2lnx+2x+3 f '' ( x ) = 2 ln x + 2 x + 3 Skip to navigation
To find the derivative f'(x) of f(x), we can use the product rule. The product rule states that if we have a function g(x) multiplied by another function h(x), the derivative of the product g(x) * h'(x) is given by g'(x)h(x) + g(x)h'(x).
In this case, we have g(x) = x^2 and h(x) = ln(x) - x. Taking the derivative of g(x) and h(x), we have:
g'(x) = 2x
h'(x) = (1/x) - 1
Now we can use the product rule to find f'(x):
f'(x) = g'(x)h(x) + g(x)h'(x)
= (2x)(ln(x) - x) + (x^2)((1/x) - 1)
= 2xln(x) - 2x^2 + x - x^2
= 2xln(x) - x^2 + x
So the correct answer is f'(x) = 2xln(x) - x^2 + x.
To find f''(x), we need to take the derivative of f'(x):
f''(x) = d/dx (2xln(x) - x^2 + x)
= 2(ln(x) + 1) - 2x + 1
= 2ln(x) + 2 - 2x + 1
= 2ln(x) - 2x + 3
So the correct answer is f''(x) = 2ln(x) - 2x + 3.