Question

suppose that $f(\frac{\pi}{4})=-7$ and $f(\frac{\pi}{4}) = 6$, and let $g(x)=f(x)\sin x$ and $h(x)=\frac{\cos x}{f(x)}$. compute each of the following. a. $g(\frac{\pi}{4})=$ b. $h(\frac{\pi}{4})=$

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 \(g^{\prime}(x)\):

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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 \(h^{\prime}(x)\):

$$
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Answer:

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