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Question: How do you use the definition of a derivative to show that if \( f\left( x \right)=\dfrac{1}{x} \) t...

How do you use the definition of a derivative to show that if f(x)=1xf\left( x \right)=\dfrac{1}{x} then f′(x)=−1x2f'\left( x \right)=-\dfrac{1}{{{x}^{2}}} ?

Explanation

Solution

start the solution with f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} part of the definition of derivative without considering the lim⁡h→0\displaystyle \lim_{h \to 0} . Then find f(x+h) from f(x) given in the question. Put the values of f(x+h) and f(x) in the definition and simplify it. Once it is simplified then proceed with the f′(x)f'\left( x \right) part and put the value of h=0 when required.

Complete step by step answer:
Definition of derivative: We know the derivative of f(x)f\left( x \right) is the function f′(x)f'\left( x \right) and is defined as f′(x)=lim⁡h→0f(x+h)−f(x)hf'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
For the proof given in the question, let’s begin with the f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} part first.
We have f(x)=1xf\left( x \right)=\dfrac{1}{x}
Then f(x+h)=1x+hf\left( x+h \right)=\dfrac{1}{x+h}
So,
f(x+h)−f(x)h =(1x+h−1x)h \begin{aligned} & \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\\ & =\dfrac{\left( \dfrac{1}{x+h}-\dfrac{1}{x} \right)}{h} \\\ \end{aligned}
Taking L.C.M. of the numerator, we get
=(1⋅x−1.(x+h)x(x+h))h =(x−x−hx(x+h))h =(−hx(x+h))h =−hx(x+h)×1h \begin{aligned} & \text{=}\dfrac{\left( \dfrac{1\cdot x-1.\left( x+h \right)}{x\left( x+h \right)} \right)}{h} \\\ & \text{=}\dfrac{\left( \dfrac{x-x-h}{x\left( x+h \right)} \right)}{h} \\\ & \text{=}\dfrac{\left( \dfrac{-h}{x\left( x+h \right)} \right)}{h} \\\ & \text{=}\dfrac{-h}{x\left( x+h \right)}\times \dfrac{1}{h} \\\ \end{aligned}
After ‘h’ got cancel out from numerator and denominator, we get
=−1x(x+h)\text{=}-\dfrac{1}{x\left( x+h \right)}
Now, Let’s come back to f′(x)f'\left( x \right) part.
f′(x)=lim⁡h→0f(x+h)−f(x)hf'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
Since we got f(x+h)−f(x)h=−1x(x+h)\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\text{=}-\dfrac{1}{x\left( x+h \right)}
So by replacing the value of f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} above, we get
f′(x)=lim⁡h→0−1x(x+h)f'\left( x \right)\text{=}\displaystyle \lim_{h \to 0}-\dfrac{1}{x\left( x+h \right)}
Here we can replace the value of h=0 as ‘h’ tends to 0
⇒f′(x)=−1x(x+0) ⇒f′(x)=−1x⋅x ⇒f′(x)=−1x2 \begin{aligned} & \Rightarrow f'\left( x \right)\text{=}-\dfrac{1}{x\left( x+0 \right)} \\\ & \Rightarrow f'\left( x \right)\text{=}-\dfrac{1}{x\cdot x} \\\ & \Rightarrow f'\left( x \right)\text{=}-\dfrac{1}{{{x}^{2}}} \\\ \end{aligned}
Hence Proved.

Note:
The solution should start with f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} part and not with f′(x)f'\left( x \right) part because if we start with the f′(x)f'\left( x \right) part there is a ‘h’ in denominator and since h→0h \to 0 , f′(x)f'\left( x \right) will be infinity. Hence we will not get any solutions. So it’s advised to proceed with the f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} part first. Else we may proceed directly with the f′(x)f'\left( x \right) part by not putting the value of ‘h’ till the f(x+h)−f(x)h\dfrac{f\left( x+h \right)-f\left( x \right)}{h} part not get simplified. The ideal way is to solve by taking parts to avoid confusion.