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Question: What is the difference between \(delta\left( x \right)\) and \(dx\) ?...

What is the difference between delta(x)delta\left( x \right) and dxdx ?

Explanation

Solution

Here in this question we have been asked to write the difference between delta(x)delta\left( x \right) and dxdx. For answering this question we will be defining the terms and will give some examples and list out the differences.

Complete step-by-step solution:
Now considering from the question we have been asked to write the difference between delta(x)delta\left( x \right) and dxdx.
From the basic concepts we know that, the delta(x)=δxdelta\left( x \right)=\delta x is generally used to represent a small increment in the value of xx whereas the dxdx is defined from the definition of the derivative limδx0f(x+δx)f(x)δx=ddxf(x)\displaystyle \lim_{\delta x \to 0}\dfrac{f\left( x+\delta x \right)-f\left( x \right)}{\delta x}=\dfrac{d}{dx}f\left( x \right) where dxdx is actually a part of the operator.
Let us consider an example f(x)=x2f\left( x \right)={{x}^{2}} the derivative of this function will be given as ddxf(x)limh0(x+h)2x2h =limh0(2x+h)2x \begin{aligned} & \dfrac{d}{dx}f\left( x \right)\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h} \\\ & =\displaystyle \lim_{h \to 0}\left( 2x+h \right)\Rightarrow 2x \\\ \end{aligned}
where hh is the small difference of xx that is δx\delta x .
Let us consider another example where the value of xx is 12 initially and it changes to 12.000112.0001 which is almost ignorable change so the δx\delta x is given as 0.00010.0001 .
Therefore we can conclude that the delta(x)delta\left( x \right) is the small change in the quantity xx whereas dxdx is an operator used in integrations and differentiations.

Note: While answering questions of this type we should be sure with the concepts that we are going to use in between the steps because this is purely a concept based question and can be answered directly. Mistakes are possible only if we misunderstand the concepts. So through practice of the concepts can help in answering questions of this type.