how can I prove that d/dx (1/x) = -1/x^2. ??? how can I prove that d/dx (1/x) = -1/x^2. ??? @MIT 18.01 Single …

Question

anonymous · Answer

1/x <=> x^-1 

d?dx x^-1 = -1x^-2, = -1/x^2 qed

anonymous · Answer

I will prove that using the definition in the lecture.
Question: Prove that $$d/dx (1/x) = -1/ x^{2}$$Answer:
$$f(x)=1/x$$$$d/dx(1/x)=\lim_{\Delta x \rightarrow 0} (f(x _{0}+\Delta x)-f(x _{0}))/\Delta x$$(Notice that we have not been able to substitute Delta x=0 in)$$=\lim_{\Delta x \rightarrow 0}(1/(x_{0}+\Delta x)-1/x_{0})/\Delta x$$$$= \lim_{\Delta x \rightarrow 0}(x_{0}-(x_{0}+\Delta x))/((x_{0}+\Delta x).x_{0}.\Delta x)$$$$=\lim_{\Delta x \rightarrow 0}(-\Delta x)/((x_{0}+\Delta x).x_{0}.\Delta x)$$(Now we cancel out Delta x)$$=\lim_{\Delta x \rightarrow 0}(-1)/((x_{0}+\Delta x).x_{0})$$(Now, we can substitute Delta x=0)$$=-1/x ^{2}$$