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Question: How do you use limit definition to find the derivative of \[f\left( x \right) = \dfrac{1}{{{x^2}}}\]...

How do you use limit definition to find the derivative of f(x)=1x2f\left( x \right) = \dfrac{1}{{{x^2}}} ?

Explanation

Solution

Hint : The given question wants us to evaluate the derivative of a given function using the first principle of derivative. The first principle of derivative involves the concepts of limits, derivatives, continuity and differentiability. It helps us to determine the derivative of the function using concepts of limits.

Complete step by step solution:
Given the functionf(x)=1x2f\left( x \right) = \dfrac{1}{{{x^2}}}, we have to calculate the derivative of the function using the first principle of derivatives.
Using formula of first principle of derivatives, we know that derivative of a function is calculated as:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
So, derivative of the given functionf(x)=1x2f\left( x \right) = \dfrac{1}{{{x^2}}},
== f′(x)=lim⁡h→0(1(x+h)2)−(1x2)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{1}{{{{\left( {x + h} \right)}^2}}}} \right) - \left( {\dfrac{1}{{{x^2}}}} \right)}}{h}
Simplifying by opening bracket,
== f′(x)=lim⁡h→0(x2−(x+h)2(x+h)2x2)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{{{x^2} - {{\left( {x + h} \right)}^2}}}{{{{\left( {x + h} \right)}^2}{x^2}}}} \right)}}{h}
Opening the whole square brackets and simplifying the expression, we get,
== f′(x)=lim⁡h→0x2−(x2+2xh+h2)h(x+h)2x2f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^2} - \left( {{x^2} + 2xh + {h^2}} \right)}}{{h{{\left( {x + h} \right)}^2}{x^2}}}
== f′(x)=lim⁡h→0x2−x2−2xh−h2h(x+h)2x2f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^2} - {x^2} - 2xh - {h^2}}}{{h{{\left( {x + h} \right)}^2}{x^2}}}
Cancelling the like terms with opposite signs, we get,
== f′(x)=lim⁡h→0−2xh−h2h(x+h)2x2f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2xh - {h^2}}}{{h{{\left( {x + h} \right)}^2}{x^2}}}
Cancelling the common terms in numerator and denominator, we get,
== f′(x)=lim⁡h→0−2x−h(x+h)2x2f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2x - h}}{{{{\left( {x + h} \right)}^2}{x^2}}}
On applying limits, we get the derivative of the given function as,
== f′(x)=lim⁡h→0−2x−0(x+0)2x2f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2x - 0}}{{{{\left( {x + 0} \right)}^2}{x^2}}}
== f′(x)=−2xx4f'(x) = \dfrac{{ - 2x}}{{{x^4}}}
== f′(x)=−2x3f'(x) = \dfrac{{ - 2}}{{{x^3}}}
Hence, the derivative of f(x)=1x2f\left( x \right) = \dfrac{1}{{{x^2}}} is f′(x)=−2x3f'(x) = \dfrac{{ - 2}}{{{x^3}}}.
So, the correct answer is “f′(x)=−2x3f'(x) = \dfrac{{ - 2}}{{{x^3}}}”.

Note : There are various methods of finding the derivatives of the given functions. The calculation of derivative using the first principle of derivative is the most basic and core method to find the derivative of a specific function. Another method of finding derivative of a function can be simply differentiating that particular function with respect to the variable or unknown using some basic differentiation rules like addition rule, subtraction rule, product rule, and quotient rule. This method is comparatively easier as it does not involve concepts limits and continuity.