For the function \(y=\sqrt{x}\) the value of \(dy\) and \(\Delta y\) are as follows:
\(\fbox{\begin\\\ \math dy=\dfrac{1}{2}\ \text{and}\ \Delta y=0.414\\\end{minispace}}\)
Further explanation:
In the question it is given that the function is \(y=\math radical(x)\). The term radical represents the square root symbol \(\left(\sqrt{}\right)\).
The function is expressed as follows:
\(y=\sqrt{x}\)
The above function is also represented as follows:
\(f(x)=\sqrt{x}\)
It is given that the value of \(x\) is \(1\) and the value of \(\Delta x\) is \(1\).
Consider that for small change in the value of \(x\) as \(\Delta x\) there occur a small change in \(y\) as \(\Delta y\).
The change in the dependent variable \(y\) is expressed as follows:
\(\fbox{\begin\\\ \Delta y=f(x+\Delta x)-f(x)\\\end{minispace}}\)
The change in the dependent variable \(y\) with respect to change in independent variable is expressed as follows:
\(\fbox{\begin\\\ \dfrac{\Delta y}{\Delta x}=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\\end{minispace}}\)
Substitute \(f(x)=\sqrt{x}\) and \(f(x+\Delta x)=\sqrt{x+\Delta x}\) in the above equation.
\(\dfrac{\Delta y}{\Delta x}=\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\)
Rationalize the above expression by multiplying and dividing the term \(\sqrt{x+\Delta x}+\sqrt{x}\).
\(\begin{aligned}\dfrac{\Delta y}{\Delta x}&=\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\\&=\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\times\dfrac{\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}\\&=\dfrac{x+\Delta x-x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}\\&=\dfrac{1}{\sqrt{x+\Delta x}+\sqrt{x}}\end{aligned}\)
Substitute \(1\) for \(x\) and \(1\) for \(\Delta x\) in the above equation.
\(\begin{aligned}\dfrac{\Delta y}{1}&=\dfrac{1}{(\sqrt{1+1}+\sqrt{1})}\\\Delta y&=\dfrac{1}{\sqrt{2}+1}\end{aligned}\)
Rationalize the above expression to obtain the value of \(\Delta y\).
\(\begin{aligned}\Delta y&=\dfrac{1}{\sqrt{2}+1}\\&=\dfrac{1}{\sqrt{2}+1}\times\dfrac{\sqrt{2}-1}{\sqrt{2}-1}\\&=\sqrt{2}-1\\&=1.414-1\\&=0.414\end{aligned}\)
Therefore, the value of \(\Delta y\) is \(0.414\).
For an infinitesimally small change in \(x\) i.e., as \(\Delta x\rightarrow0\) then the equation (1) is expressed as follows:
\(\fbox{\begin\\\ \dfrac{dy}{dx}=\lim_{x\to0}\dfrac{(f(x+\Delta x)+f(x))}{\Delta x}\\\end{minispace}}\)
Substitute \(f(x)=\sqrt{x}\) and \(f(x+\Delta x)=\sqrt{x+\Delta x}\) in the above equation.
\(\dfrac{dy}{dx}=\lim_{x\to0}\left(\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\right)\)
Rationalize the above expression as follows:
\(\begin{aligned}\dfrac{dy}{dx}&=\lim_{x\to0}\left(\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\times\dfrac{\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}\right)\\&=\lim_{x\to0}\left(\dfrac{x+\Delta x-x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}\right)\\&=\lim_{x\to0}\left(\dfrac{1}{\sqrt{x+\Delta x}+\Delta x}\right)\\&=\dfrac{1}{\sqrt{\Delta x}+\Delta x}\end{aligned}\)
Substitute \(1\) for \(\Delta x\) and \(1\) for \(dx\)in the above equation.
\(\begin{aligned}\dfrac{dy}{1}&=\dfrac{1}{\srqt{1}+1}\\dy&=\dfrac{1}{2}\end{aligned}\)
Therefore, the value of \(dy\) is \(1\).
Thus, for the function \(y=\sqrt{x}\) the value of \(dy\) and \(\Delta y\) are as follows:
\(\fbox{\begin\\\ \math dy=\dfrac{1}{2}\ \text{and}\ \Delta y=0.414\\\end{minispace}}\)
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Answer details:
Grade: Senior school
Subject: Mathematics
Chapter: Curve sketching
Keywords: Curve, graph, radical, quadratic, expression, roots, y=rootx, delta y, dy, derivative, dx, delta x, round off, decmials, decimal places, rationalize.
