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Derivative at a point via the limit definition

Let f(x)=1xf(x) = \dfrac{1}{x}.

Use the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, to compute f′(2)f'(2) for f(x)=1xf(x) = \dfrac{1}{x}. Show the steps of setting up the difference quotient, simplifying it algebraically, and then evaluating the limit as h→0h \to 0. What is f′(2)f'(2)?

Hint

Write out f(2+h)=12+hf(2+h) = \dfrac{1}{2+h} and f(2)=12f(2) = \dfrac{1}{2}, combine the resulting fractions in the numerator over a common denominator, then simplify the compound fraction before canceling the hh in the denominator.

Step-by-step solution

Start with the difference quotient: f(2+h)−f(2)h=12+h−12h\dfrac{f(2+h)-f(2)}{h} = \dfrac{\frac{1}{2+h} - \frac{1}{2}}{h}. Combine the two fractions in the numerator using the common denominator 2(2+h)2(2+h), giving 2−(2+h)2(2+h)=−h2(2+h)\dfrac{2 - (2+h)}{2(2+h)} = \dfrac{-h}{2(2+h)}.

Substitute this back into the difference quotient to get −h2h(2+h)\dfrac{-h}{2h(2+h)}, and cancel the common factor of hh (valid since h≠0h \neq 0 in the limit) to obtain −12(2+h)\dfrac{-1}{2(2+h)}. Taking the limit as h→0h \to 0 gives −12(2)=−14\dfrac{-1}{2(2)} = -\dfrac{1}{4}, so f′(2)=−14f'(2) = -\dfrac{1}{4}.

Sign of the derivative value at the computed point

Hint
Step-by-step solution

Negative. The computation in part a showed that f prime of two equals negative one fourth, which is a negative number. This makes sense since one over x is decreasing at x equals two, so its instantaneous rate of change there should be negative.

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