For the function [tex]y=\sqrt{x}[/tex] the value of [tex]dy[/tex] and [tex]\Delta y[/tex] are as follows:
[tex]\fbox{\begin\\\ \math dy=\dfrac{1}{2}\ \text{and}\ \Delta y=0.414\\\end{minispace}}[/tex]
Further explanation:
In the question it is given that the function is [tex]y=\math radical(x)[/tex]. The term radical represents the square root symbol [tex]\left(\sqrt{}\right)[/tex].
The function is expressed as follows:
[tex]y=\sqrt{x}[/tex]
The above function is also represented as follows:
[tex]f(x)=\sqrt{x}[/tex]
It is given that the value of [tex]x[/tex] is [tex]1[/tex] and the value of [tex]\Delta x[/tex] is [tex]1[/tex].
Consider that for small change in the value of [tex]x[/tex] as [tex]\Delta x[/tex] there occur a small change in [tex]y[/tex] as [tex]\Delta y[/tex].
The change in the dependent variable [tex]y[/tex] is expressed as follows:
[tex]\fbox{\begin\\\ \Delta y=f(x+\Delta x)-f(x)\\\end{minispace}}[/tex]
The change in the dependent variable [tex]y[/tex] with respect to change in independent variable is expressed as follows:
[tex]\fbox{\begin\\\ \dfrac{\Delta y}{\Delta x}=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\\end{minispace}}[/tex]
Substitute [tex]f(x)=\sqrt{x}[/tex] and [tex]f(x+\Delta x)=\sqrt{x+\Delta x}[/tex] in the above equation.
[tex]\dfrac{\Delta y}{\Delta x}=\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}[/tex]
Rationalize the above expression by multiplying and dividing the term [tex]\sqrt{x+\Delta x}+\sqrt{x}[/tex].
[tex]\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}[/tex]
Substitute [tex]1[/tex] for [tex]x[/tex] and [tex]1[/tex] for [tex]\Delta x[/tex] in the above equation.
[tex]\begin{aligned}\dfrac{\Delta y}{1}&=\dfrac{1}{(\sqrt{1+1}+\sqrt{1})}\\\Delta y&=\dfrac{1}{\sqrt{2}+1}\end{aligned}[/tex]
Rationalize the above expression to obtain the value of [tex]\Delta y[/tex].
[tex]\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}[/tex]
Therefore, the value of [tex]\Delta y[/tex] is [tex]0.414[/tex].
For an infinitesimally small change in [tex]x[/tex] i.e., as [tex]\Delta x\rightarrow0[/tex] then the equation (1) is expressed as follows:
[tex]\fbox{\begin\\\ \dfrac{dy}{dx}=\lim_{x\to0}\dfrac{(f(x+\Delta x)+f(x))}{\Delta x}\\\end{minispace}}[/tex]
Substitute [tex]f(x)=\sqrt{x}[/tex] and [tex]f(x+\Delta x)=\sqrt{x+\Delta x}[/tex] in the above equation.
[tex]\dfrac{dy}{dx}=\lim_{x\to0}\left(\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\right)[/tex]
Rationalize the above expression as follows:
[tex]\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}[/tex]
Substitute [tex]1[/tex] for [tex]\Delta x[/tex] and [tex]1[/tex] for [tex]dx[/tex]in the above equation.
[tex]\begin{aligned}\dfrac{dy}{1}&=\dfrac{1}{\srqt{1}+1}\\dy&=\dfrac{1}{2}\end{aligned}[/tex]
Therefore, the value of [tex]dy[/tex] is [tex]1[/tex].
Thus, for the function [tex]y=\sqrt{x}[/tex] the value of [tex]dy[/tex] and [tex]\Delta y[/tex] are as follows:
[tex]\fbox{\begin\\\ \math dy=\dfrac{1}{2}\ \text{and}\ \Delta y=0.414\\\end{minispace}}[/tex]
Learn more:
1. A problem to determine the equation of the line https://brainly.com/question/5337932
2. A problem to determine the point slope form of line https://brainly.com/question/4097107
3. Aproblem on lines https://brainly.com/question/1646698
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.
