suppose that $f(\\frac{\\pi}{4})=-7$ and $f(\\frac{\\pi}{4}) = 6$, and let $g(x)=f(x)\\sin x$ and…

Explanation:

Step1: Find the derivative of (g(x)) using the product - rule

The product - rule states that if (g(x)=u(x)v(x)), then (g^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)). Here, (u = f(x)) and (v=\sin x), so (g^{\prime}(x)=f^{\prime}(x)\sin x + f(x)\cos x).

Step2: Evaluate (g^{\prime}(\frac{\pi}{4}))

Substitute (x = \frac{\pi}{4}) into (g^{\prime}(x)): [ \begin{align*} g^{\prime}(\frac{\pi}{4})&=f^{\prime}(\frac{\pi}{4})\sin(\frac{\pi}{4})+f(\frac{\pi}{4})\cos(\frac{\pi}{4})\ &=6\times\frac{\sqrt{2}}{2}+(- 7)\times\frac{\sqrt{2}}{2}\ &=\frac{6\sqrt{2}-7\sqrt{2}}{2}\ &=-\frac{\sqrt{2}}{2} \end{align*} ]

Step3: Find the derivative of (h(x)) using the quotient - rule

The quotient - rule states that if (h(x)=\frac{u(x)}{v(x)}), then (h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}). Here, (u(x)=\cos x), (v(x)=f(x)), so (h^{\prime}(x)=\frac{-\sin x\cdot f(x)-\cos x\cdot f^{\prime}(x)}{f^{2}(x)}).

Step4: Evaluate (h^{\prime}(\frac{\pi}{4}))

Substitute (x = \frac{\pi}{4}) into (h^{\prime}(x)): [ \begin{align*} h^{\prime}(\frac{\pi}{4})&=\frac{-\sin(\frac{\pi}{4})\cdot f(\frac{\pi}{4})-\cos(\frac{\pi}{4})\cdot f^{\prime}(\frac{\pi}{4})}{f^{2}(\frac{\pi}{4})}\ &=\frac{-\frac{\sqrt{2}}{2}\times(-7)-\frac{\sqrt{2}}{2}\times6}{(-7)^{2}}\ &=\frac{\frac{7\sqrt{2}}{2}- \frac{6\sqrt{2}}{2}}{49}\ &=\frac{\frac{\sqrt{2}}{2}}{49}\ &=\frac{\sqrt{2}}{98} \end{align*} ]

Answer:

a. (g^{\prime}(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}) b. (h^{\prime}(\frac{\pi}{4})=\frac{\sqrt{2}}{98})