Question

j(x) = \frac{3}{x^2 + 1}

Explanation:

Assuming the problem is to find the derivative of \( J(x)=\frac{3}{x^{2}+1} \), we use the quotient rule or rewrite it as a power function and use the chain rule. Here we use the chain rule by rewriting \( J(x) = 3(x^{2}+1)^{-1} \).

Step 1: Identify the outer and inner functions

Let \( u = x^{2}+1 \), so \( J(u)=3u^{-1} \). The derivative of the outer function with respect to \( u \) is \( J^\prime(u)= - 3u^{-2} \), and the derivative of the inner function with respect to \( x \) is \( u^\prime = 2x \).

Step 2: Apply the chain rule

The chain rule states that \( J^\prime(x)=J^\prime(u)\cdot u^\prime \). Substituting the values we found:
\( J^\prime(x)=-3u^{-2}\cdot(2x) \)
Substitute back \( u = x^{2}+1 \):
\( J^\prime(x)=-3(x^{2}+1)^{-2}\cdot(2x)=-\frac{6x}{(x^{2}+1)^{2}} \)

Answer:

The derivative of \( J(x)=\frac{3}{x^{2}+1} \) is \( \boldsymbol{-\frac{6x}{(x^{2}+1)^{2}}} \) (if the problem was to find the derivative; if it was a different operation, the solution would change accordingly, but since the problem was not fully specified, this is a common operation for this function).