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Worked solution

Prove the reciprocal derivative from first principles

Let f:x↦1/xf:x\mapsto1/x. Using the first principles of differentiation, show that f′(x)=−1/x2f^{\prime}(x)=-1/x^2.

CJC · Lecture appendix · Example A

Combine the difference quotient before cancelling

Fix x≠0x\ne0. For h≠0h\ne0 and x+h≠0x+h\ne0, [f(x+h)−f(x)]/h=[1/(x+h)−1/x]/h=[x−(x+h)]/[hx(x+h)]=−1/[x(x+h)]=−1/(x2+xh)[f(x+h)-f(x)]/h=[1/(x+h)-1/x]/h=[x-(x+h)]/[hx(x+h)]=-1/[x(x+h)]=-1/(x^2+xh).

How did this part go?

Take the limit at a valid original input

As h→0h\to0, the denominator tends to x2≠0x^2\ne0. Thus f′(x)=lim⁡h→0[−1/(x2+xh)]=−1/x2f^{\prime}(x)=\lim_{h\to0}[-1/(x^2+xh)]=-1/x^2, as printed. The original function is undefined at x=0x=0; the proof does not assign it a derivative there.

How did this part go?

Common mistakes

Cancel a nonzero increment before taking its limit, and keep x≠0.

Solution notes

Both originals print the complete appendix working. Source increments are written δx; this lesson uses h. No printed marks or past-paper attribution. Domain qualifications are independently authored. Appendix letters remain in the source label and identity; internal number 0 does not renumber them as printed numbered examples.

Read the hint