homework3: problem 29\n(1 point)\nif\n f(x)=\frac{6 - x^{2}}{7 + x^{2}} \nfind ( f^{prime}(x) ).\n(…

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here, (u = 6 - x^{2}), so (u^\prime=-2x); (v = 7 + x^{2}), so (v^\prime = 2x). [ \begin{align*} f^\prime(x)&=\frac{(-2x)(7 + x^{2})-(6 - x^{2})(2x)}{(7 + x^{2})^{2}}\ \end{align*} ]

Step2: Expand the numerator

Expand ((-2x)(7 + x^{2})=-14x-2x^{3}) and ((6 - x^{2})(2x)=12x - 2x^{3}). [ \begin{align*} f^\prime(x)&=\frac{-14x-2x^{3}-(12x - 2x^{3})}{(7 + x^{2})^{2}}\ &=\frac{-14x-2x^{3}-12x + 2x^{3}}{(7 + x^{2})^{2}} \end{align*} ]

Step3: Simplify the numerator

Combine like - terms in the numerator: (-14x-12x=-26x), and (-2x^{3}+2x^{3}=0). So (f^\prime(x)=\frac{-26x}{(7 + x^{2})^{2}})

Step4: Find (f^\prime(4))

Substitute (x = 4) into (f^\prime(x)). [ \begin{align*} f^\prime(4)&=\frac{-26\times4}{(7 + 4^{2})^{2}}\ &=\frac{-104}{(7 + 16)^{2}}\ &=\frac{-104}{23^{2}}\ &=\frac{-104}{529} \end{align*} ]

Answer:

(f^\prime(x)=\frac{-26x}{(7 + x^{2})^{2}}), (f^\prime(4)=\frac{-104}{529})