Question

this is section 3.1 problem 40: for y=(x^2 - 3)^3, when x = 2 and δx=dx = 0.01: δy=, dy=. round to three decimal places unless the exact answer has less decimal places. hint: follow example 5. resources ebook

Explanation:

Step1: Find the derivative of $y=(x^{2}-3)^{3}$

Use the chain - rule. Let $u = x^{2}-3$, then $y = u^{3}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=3u^{2}$, and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=2x$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3(x^{2}-3)^{2}\cdot2x = 6x(x^{2}-3)^{2}$.

Step2: Calculate $dy$

We know that $dy=\frac{dy}{dx}dx$. Given $x = 2$ and $dx=\Delta x=0.01$. First, find $\frac{dy}{dx}$ at $x = 2$: $\frac{dy}{dx}\big|_{x = 2}=6\times2\times(2^{2}-3)^{2}=12\times1^{2}=12$. Then $dy=\frac{dy}{dx}dx=12\times0.01 = 0.120$.

Step3: Calculate $\Delta y$

$\Delta y=f(x + \Delta x)-f(x)$. Here, $x = 2$, $\Delta x=0.01$, so $x+\Delta x=2 + 0.01=2.01$. $f(x)=(x^{2}-3)^{3}$, then $f(2)=(2^{2}-3)^{3}=1^{3}=1$, and $f(2.01)=((2.01)^{2}-3)^{3}=(4.0401 - 3)^{3}=(1.0401)^{3}=1.0401\times1.0401\times1.0401\approx1.125$. So $\Delta y=f(2.01)-f(2)\approx1.125 - 1=0.125$.

Answer:

$dy = 0.120$, $\Delta y=0.125$