We find the derivative using the Product Rule, which says that if f(x) = uv, then f′(x) = u + v. Here f(x) = (x^4 − x^2 )(2x^3 + x), so u = x^4 − x^2 and v = 2x^3 + x. Using the Product Rule, we
get f′(x) = (x^4 − x^2 )(6x^2 + 1) + (2x^3 + x)(4x^3 − 2x). Now we don’t simplify. We simply plug in x= 1 to get f′(x) = ((1)^4 − (1)^2 )(6(1)^2 + 1) + (2(1)^3 + (1))(4(1)^3 − 2(1)) = 6.14.
We find the derivative using the Quotient Rule, which says that if f(x) = , then f′(x) = . Here f(x) = , so u = x^2 + 2x and v = x^4 − x^3 . Using the Quotient Rule, we
get f′(x) = . Now, we don’t simplify. We simply plug in x = 2to get. f′(x) = = −.15.
We find the derivative using the Chain Rule, which says that if y = f(g(x)), then y′ = . Here f(x) = = (x^4 + x^2 ). Using the Chain Rule, we get f′(x) = (x^4
+ x^2 ) (4x^3 + 2x). This can be simplified to f′(x) = .16.
We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here = 2u and = −1(x − 1)−2 = − . Thus, = (2u) and
because u = , = = −.17. −24
We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then