Question

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

Answer 1

Hint: start the solution with $ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} $ part of the definition of derivative without considering the $ \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'\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\left( x \right) $ is the function $ f'\left( x \right) $ and is defined as $ f'\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 $ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} $ part first.
We have $ f\left( x \right)=\dfrac{1}{x} $
Then $ f\left( x+h \right)=\dfrac{1}{x+h} $
So,
 $ \begin{align}
  & \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
 & =\dfrac{\left( \dfrac{1}{x+h}-\dfrac{1}{x} \right)}{h} \\
\end{align} $
Taking L.C.M. of the numerator, we get
  $ \begin{align}
  & \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{align} $
After ‘h’ got cancel out from numerator and denominator, we get
 $ \text{=}-\dfrac{1}{x\left( x+h \right)} $
Now, Let’s come back to $ f'\left( x \right) $ part.
 $ f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} $
Since we got $ \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 $ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} $ above, we get
 $ 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
 $ \begin{align}
  & \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{align} $
Hence Proved.

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