Question

Find the derivative of the following function from first principle $\dfrac{1}{{{x^2}}}$

Answer 1

We have to find the derivative of the given function using the first principle method . We solve this question using the concept of limits and derivatives . We should have the knowledge of the formula of the first principle method . First we will substitute the values of the given function in the formula and then we will solve the obtained expression for the required limit , such that on solving the limit we obtain the value of the derivative of the given function .

Complete step-by-step solution:
Given :
The formula of first principle of derivatives is given as :
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {f\left( {x + h} \right) - f\left( x \right)} \right]}}{h}$
Let us consider that the given function be $f\left( x \right)$ such that :
$f\left( x \right) = \dfrac{1}{{{x^2}}}$
Also , on substituting the value of $x$ by $x + h$ the function becomes as :
$f\left( {x + h} \right) = \dfrac{1}{{{{\left( {x + h} \right)}^2}}}$
Now , on substituting the value of $f\left( x \right)$ and $f\left( {x + h} \right)$ in the formula of first principle , we get
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {\dfrac{1}{{{{\left( {x + h} \right)}^2}}} - \dfrac{1}{{{x^2}}}} \right]}}{h}$
Solving the above expression by taking L.C.M. , we get
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {{x^2} - {{\left( {x + h} \right)}^2}} \right]}}{{h{{\left( {x + h} \right)}^2}{x^2}}}$
Now , we will expand the term using the formula of square of sum of two numbers which is given as :
${(a + b)^2} = {a^2} + {b^2} + 2ab$
Using the formula , we get the expression as :
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {{x^2} - \left( {{x^2} + {h^2} + 2xh} \right)} \right]}}{{h{{\left( {x + h} \right)}^2}{x^2}}}$
On further solving , we get
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ { - {h^2} - 2hx} \right]}}{{h{{\left( {x + h} \right)}^2}{x^2}}}$
Cancelling h from both the numerator and denominator , we get
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ { - h - 2x} \right]}}{{{{\left( {x + h} \right)}^2}{x^2}}}$
Now , we will put the value of the limit in the expression .
Putting the value of the limit in the expression , we get
$f'\left( x \right) = \dfrac{{\left[ { - 0 - 2x} \right]}}{{{{\left( {x + 0} \right)}^2}{x^2}}}$
$f'\left( x \right) = \dfrac{{ - 2x}}{{{x^2} \times {x^2}}}$
Cancelling x from both the numerator and denominator , we get
$f'\left( x \right) = \dfrac{{ - 2}}{{{x^2} \times x}}$
On further simplifying , we get
$f'\left( x \right) = \dfrac{{ - 2}}{{{x^3}}}$
Hence , the value of the derivative of the given function $\dfrac{1}{{{x^2}}}$ using the first principle method is $\dfrac{-2}{{{x^3}}}$ .

Note: We can also check our answer of the derivative obtained using the formula of the first principle method by comparing it with the result which we will obtain directly using the formulas of derivatives of the function . Differentiating the function using any of the one method either using the first principle method or by directly using the various formulas of derivatives we should obtain the same value of derivative.