Question

Assertion \((\mathbf{A}):\) Let \(g(x)=f(x) \sin x\), where \(f(x)\) is a twice differentiable function on \((-\infty, \infty)\) such that \(\mathrm{f}^{\prime}(-\pi)=1\). The value of \(g^{\prime \prime}(-\pi)\) equals \(-2\). Reason \((\mathbf{R}):\) We have $g^{\prime \prime}(x)=f(x)(-\sin x)+f^{\prime}(x) \cos x+\( \)\mathrm{f}^{\prime}(\mathrm{x}) \cos x+\mathrm{f}^{\prime \prime}(\mathrm{x}) \sin \mathrm{x}$. Hence \(\mathrm{g}^{\prime \prime}(-\pi)=-2\)

Short answer

Question: Verify that the second derivative of the function \(g(x) = f(x) \sin x\) at \(x = -\pi\) is equal to -2, given that \(f'(x)\) is a twice differentiable function and \(f'(-\pi) = 1\). Solution: We found the second derivative of \(g(x)\) to be \(g''(x) = f''(x) \sin x + 2f'(x) \cos x - f'(x) \sin x\). At \(x = -\pi\) and given \(f'(-\pi) = 1\), we have \(g''(-\pi) = -2\), verifying the given information.

Step-by-step solution

Find the first derivative of \(g(x)\)

To find the first derivative of \(g(x)=f(x) \sin x\), we apply the product rule: \((uv)' = u'v + uv'\) Here, \(u = f(x)\) and \(v = \sin x\) \(g'(x) = f'(x) \sin x + f(x) \cos x\)

Find the second derivative of \(g(x)\)

Now, we will find the second derivative of \(g(x)\) by differentiating \(g'(x)\). First, we differentiate \(f'(x) \sin x\): \((f'(x) \sin x)' = f''(x) \sin x + f'(x) \cos x\) Next, we differentiate \(f(x) \cos x\): \((f(x) \cos x)' = f'(x)(-\sin x) + f''(x) \cos x\) Now, we add both the above derivatives together: \(g''(x) = f''(x) \sin x + f'(x) \cos x + f'(x)(-\sin x) + f''(x) \cos x\) Simplifying, we get: \(g''(x) = f''(x) \sin x + 2f'(x) \cos x - f'(x) \sin x\)

Use given information to find the value of \(g''(-\pi)\)

We are given that \(f'(-\pi) = 1\) and \(g''(-\pi) = -2\). We will use this information to find the value of \(g''(-\pi)\). \(g''(-\pi) = f''(-\pi) \sin (-\pi) + 2f'(-\pi) \cos (-\pi) - f'(-\pi) \sin (-\pi)\) Since \(f'(-\pi) = 1\), \(g''(-\pi) = f''(-\pi) \sin (-\pi) + 2(1) \cos (-\pi) - 1 \sin (-\pi)\) We know that \(\sin (-\pi) = 0\) and \(\cos (-\pi) = -1\), so \(g''(-\pi) = f''(-\pi)(0) + 2(-1) - 1 (0)\) \(g''(-\pi) = -2\) Thus, we have verified that the value of \(g''(-\pi) = -2\).