Question

Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) y = radical(x) x = 1, Δx = 1 find Δy and dy.

Answer 1

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.

Answer 2

Base on the data you have given and with my calculation, i think the best value for Delta X and Dy is delta x = 2 and the DY = 1/2.I hope you are satisfied with my answer and feel free to ask for more if you have more clarifications and further questions