Question

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=\frac{x^{2}}{x^{2}-4} $$

Short answer

The function is differentiable for all \(x\) except \(x=-2\) and \(x=2\).

Step-by-step solution

Simplify the Function

Simplify the function if possible. However, in this case, the function \(y=\frac{x^{2}}{x^{2}-4}\) cannot be further simplified and so this step is skipped.

Apply the Quotient Rule

Use the quotient rule of differentiation which states that if there are two differentiable functions u and v (both in terms of x), their quotient is also a differentiable function with derivative \(u'v - uv'\) over \(v^{2}\). Apply this to the provided function to find \(y'=\frac{2x*(x^{2}-4) - x^{2} * 2x}{(x^{2}-4)^{2}}\). Simplify this to get \(y'= \frac{-8x}{(x^{2}-4)^{2}}\).

Find Points of Non-Differentiability

Recall that a function is not differentiable at points where the derivative does not exist. This includes points where the denominator equals to zero. Solving the equation \(x^{2}-4=0\) gives \(x=-2, 2\). So the function is not differentiable at \(x=-2\) and \(x=2\)