a) Test the continuity of the function f(x) = cases x^2 4xx 4 & when x 4 \\ 4 & when x = 4 cases at x = 4 . b) Find dydx if y = xx^2 1 .

a) Testing the Continuity of the Function at $x=4$

To test the continuity of the function $f(x)$ at $x=4$, we need to check the following three conditions:

1. $f(x)$ is defined at $x=4$.
2. ${\lim }_{x\to 4}f(x)$ exists.
3. ${\lim }_{x\to 4}f(x)=f(4)$.

Step 1: Check if $f(x)$ is defined at $x=4$

The function is explicitly defined at $x=4$ as: $$f(4)=4$$ So, $f(x)$ is defined at $x=4$.

Step 2: Find ${\lim }_{x\to 4}f(x)$

The function is defined piecewise as: $$f(x)=\{\begin{array}{ll}\frac{x^2-4x}{x-4}& \text{when }x\ne 4\\ 4& \text{when }x=4\end{array}$$

For $x\ne 4$, we simplify the expression: $$\frac{x^2-4x}{x-4}=\frac{x(x-4)}{x-4}$$ For $x\ne 4$, the denominator $x-4\ne 0$, so we can cancel it out: $$\frac{x(x-4)}{x-4}=x$$ Thus, for $x\ne 4$, $f(x)=x$.

Now, we find the limit as $x$ approaches 4: $$\underset{x\to 4}{\lim }f(x)=\underset{x\to 4}{\lim }x=4$$

Step 3: Compare ${\lim }_{x\to 4}f(x)$ and $f(4)$

We have: $$\underset{x\to 4}{\lim }f(x)=4\text{and}f(4)=4$$ Since ${\lim }_{x\to 4}f(x)=f(4)$, the function $f(x)$ is continuous at $x=4$.

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b) Finding $\frac{dy}{dx}$ if $y=\frac{x}{x^2-1}$

We are given: $$y=\frac{x}{x^2-1}$$

To find $\frac{dy}{dx}$, we use the Quotient Rule for differentiation, which states: $$\frac{dy}{dx}=\frac{(x^2-1)\cdot \frac{d}{dx}(x)-x\cdot \frac{d}{dx}(x^2-1)}{(x^2-1{)}^2}$$

Step 1: Differentiate the numerator and denominator

• The numerator is $x$, so $\frac{d}{dx}(x)=1$.
• The denominator is $x^2-1$, so $\frac{d}{dx}(x^2-1)=2x$.

Step 2: Apply the Quotient Rule

Substitute the derivatives into the quotient rule formula: $$\frac{dy}{dx}=\frac{(x^2-1)\cdot 1-x\cdot 2x}{(x^2-1{)}^2}$$

Step 3: Simplify the expression

$$\frac{dy}{dx}=\frac{x^2-1-2x^2}{(x^2-1{)}^2}=\frac{-x^2-1}{(x^2-1{)}^2}$$ $$\frac{dy}{dx}=\frac{-(x^2+1)}{(x^2-1{)}^2}$$

Thus, the derivative is: $$\frac{dy}{dx}=\boxed{\frac{-(x^2+1)}{(x^2-1{)}^2}}$$