Question

How do you find the derivatives of\[f\left( x \right)=1-{{x}^{2}}\] using the limit process?

Answer 1

Hint: These types of problems are pretty straight forward and are very easy to solve. For problems like these we need to remember all the concepts and equations of the theory of limits including the first principle of limits to find the derivatives. According to the first principle of limits, say we have a function \[f\left( x \right)\] and we consider a point on the curve \[y=f\left( x \right)\] as \[\left( x,f\left( x \right) \right)\] and another point \[\left( x+h,f\left( x+h \right) \right)\] , where \[h\] is an infinitesimal small quantity, then the derivative of the function $f\left( x \right)$ is defined as,
$f'\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Now, substituting the value of the function \[f\left( x \right)\] in the above equation, we can easily find out its derivative.

Complete step by step answer:
Now, we start off the solution to the given problem by writing that,
We substitute the value of the function \[f\left( x \right)\] in the above equation as,
\[\begin{align}
  & {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{\left( 1-{{\left( x+h \right)}^{2}} \right)-\left( 1-{{x}^{2}} \right)}{h} \\
\end{align}\]
On breaking up the terms and evaluating we get,
\[\begin{align}
  & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{\left( 1-{{x}^{2}}-{{h}^{2}}-2hx \right)-\left( 1-{{x}^{2}} \right)}{h} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{1-{{x}^{2}}-{{h}^{2}}-2hx-1+{{x}^{2}}}{h} \\
\end{align}\]
Now, cancelling off the positive and negative same terms we get,
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{-{{h}^{2}}-2hx}{h}\]
Now, dividing both the terms by \[h\] we get,
\[\begin{align}
  & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}-\dfrac{{{h}^{2}}}{h}-\dfrac{2hx}{h} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}-h-2x \\
\end{align}\]
Now, putting the value of \[h=0\] , we get,
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=-2x\]
Thus our answer to the given problem is \[-2x\] .

Note:
For these types of problems which involve finding the derivative using the limit process, we need to first of all remember the theory of limits and all the formulae and equations involving limits, failing which solving of the problem may seem impossible. We then need to plug in the given function into the known equation that we have of the limits and then evaluate the limit normally as we do when we solve simple limit problems. We can also cross check the answer to the problem by performing normal differentiation to the given function.