get f′(x) = . This can be simplified to f′(x) = .22. 0
Here we will find the derivative using the Chain Rule. We will also need the Quotient Rule totake the derivative of the expression inside the parentheses. The Chain Rule says that if y =f(g(x)), then y′ = and the Quotient Rule says that if f(x) = , then f′(x) = . We get f′(x) = . Now we don’t simplify. We
simply plug in x = 1 to get f′(x) = = 0.23.
We find the derivative using the Quotient Rule, which says that if f(x) = , then f′(x) = . We will also need the Chain Rule to take the derivative of the expression in the
denominator. The Chain Rule says that if y = f(g(x)), then y′= . Here f(x) = , so u = x and v = (1 + x^2 )^2 . We get f(x) = . Now we don’tsimplify. We simply plug in x = 1 to get f(x) = = −.24.
We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here and . Now, we plug v = 2 into
the derivative. Note that, where v = 2, x = . We get + + and , so .- −4 csc^2 (4x)