Question

How do you find the derivative of $f\left( x \right)=7{{x}^{2}}-3$ using the limit definition?

Answer 1

Assume ‘h’ as the small change in x and hence find the small change in the function f (x) given as f(x + h). Now, use the limit definition of derivative and apply the formula: $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$, substitute the value of the given functions and simplify the limit to get the answer.

Complete step by step solution:
Here, we have been provided with the function $f\left( x \right)=7{{x}^{2}}-3$ and we are asked to find its derivative using the limit definition.
We know that derivative of a function is defined as the rate of change of function. In other words we can say that it is the measure of change in the value of the function with respect to the change in the value of the variable on which it depends. This change is infinitesimally small, that means tending to zero. Mathematically we have,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$
In the above formula we have ‘h’ as the infinitesimally small change in the variable x.
Now, let us come to the question. We have the function $f\left( x \right)=7{{x}^{2}}-3$, so substituting (x + h) in place of x in the function, we get,
$\Rightarrow f\left( x+h \right)=7{{\left( x+h \right)}^{2}}-3$
Now, substituting the value of the functions f (x) and f (x + h) in the limit formula, we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\left( 7{{\left( x+h \right)}^{2}}-3 \right)-\left( 7{{x}^{2}}-3 \right)}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{7{{\left( x+h \right)}^{2}}-3-7{{x}^{2}}+3}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{7{{\left( x+h \right)}^{2}}-7{{x}^{2}}}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{7\lim }}\,\left( \dfrac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h} \right) \\\ \end{aligned}$
Using the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{7\lim }}\,\left( \dfrac{\left( x+h-x \right)\left( x+h+x \right)}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{7\lim }}\,\left( \dfrac{\left( h \right)\left( 2x+h \right)}{h} \right) \\\ \end{aligned}$
Cancelling the common terms inside the limit we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{7\lim }}\,\left( 2x+h \right)$
Substituting the value h = 0, we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=7\left( 2x+0 \right) \\\ & \Rightarrow f'\left( x \right)=14x \\\ \end{aligned}$
Hence, the above relation is required answer

Note: One may note that the limit definition of derivative is also known as the first principle of differentiation. It is the basic definition of derivative and all the formulas of differentiation of different functions are derived from the first principle. You must remember the definition of derivative and the formula of first principle to solve the above question and do not use the formula: $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ to solve the question. You can verify the answer using this formula.