Question 1 What is wrong with the following argument claiming to prove that d 1/x = -1/x2 dx There are actually two errors, but you only need to find one. Proof: Let f(x) =1/x. Then 1 = f(x)x. It follows by Theorem 3.2.6 that d d 0 = 1 = dx dx (f(x)x) = of'(x) +f(x)- x = xf(x)+f(x) d.x It follows that of'(x) = -f(x) and, recalling the definition of f(x), yields that xf(x) = -1/x. Solving for f'(x) yields that dx 1/x = f'(x) = -1/x2