Derivatives of Mathematical Functions
The given image shows two mathematical functions, f(x) and g(x), and asks for the derivation of these functions at certain points.
The functions are:
f(x) = x^3 + 1
g(x) = x^2 - x
You are asked to do two things:
a) Find the derivative of f(x)g(x) at x = 1.
b) Find the derivative of the quotient f(x)/g(x) at x = -1.
Let's solve each part separately.
a) To find the derivative of f(x)g(x), we first need to find the product of these functions and then take the derivative of the product, and evaluate it at x = 1.
The product of f(x) and g(x) is:
f(x)g(x) = (x^3 + 1)(x^2 - x)
Instead of multiplying and then differentiating, we will use the product rule for differentiation, which states that (fg)' = f'g + fg'. Here are the derivatives of f(x) and g(x):
f'(x) = d/dx (x^3 + 1) = 3x^2
g'(x) = d/dx (x^2 - x) = 2x - 1
Now apply the product rule:
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
= (3x^2)(x^2 - x) + (x^3 + 1)(2x - 1)
Now we evaluate this derivative at x = 1:
(3(1)^2)(1^2 - 1) + (1^3 + 1)(2(1) - 1)
= 3(1)(0) + (2)(1)
= 0 + 2
= 2
So the derivative of f(x)g(x) at x = 1 is 2.
b) To find the derivative of the quotient f(x)/g(x), we will use the quotient rule for differentiation, which states that (f/g)' = (f'g - fg') / g^2.
Again, we will need to use f'(x) = 3x^2 and g'(x) = 2x - 1, and f(x) = x^3 + 1 and g(x) = x^2 - x.
Applying the quotient rule, we get:
(f(x)/g(x))' = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2
= ((3x^2)(x^2 - x) - (x^3 + 1)(2x - 1)) / (x^2 - x)^2
Now we evaluate this derivative at x = -1:
= ((3(-1)^2)((-1)^2 - (-1)) - ((-1)^3 + 1)(2(-1) - 1)) / ((-1)^2 - (-1))^2
= ((3)(1)(1 + 1) - ((-1) + 1)(-2 - 1)) / (1 + 1)^2
= (6 - (0)(-3)) / 4
= 6 / 4
= 3 / 2
So the derivative of f(x)/g(x) at x = -1 is 3/2.
