Problem 55 - Calculus of a Single Variable

Question

Find the derivative of the function. \(y=8 \arcsin \frac{x}{4}-\frac{x \sqrt{16-x^{2}}}{2}\)

Step-by-Step Solution

Answer

\(y^{'}= \frac{2}{\sqrt{16-x^{2}}}-\frac{1}{2}\sqrt{16-x^{2}}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\)

Step 1: Determine the Derivative of the sin^(-1)

We first look at the derivative of \(\arcsin u\), which is \( \frac{1}{\sqrt{1-u^{2}}}\). In our case, \(y=8 \arcsin \frac{x}{4}\), \(u=\frac{x}{4}\). Therefore, using the chain rule, we get \( \frac{8}{4\sqrt{1-(\frac{x}{4})^{2}}}\cdot u^{'}\). Now, all that is left is to find the derivative of \(u\), which is \(u^{'} = \frac{1}{4}\). So, the derivative of the first part of the function is \( \frac{2}{\sqrt{16-x^{2}}}\)

Step 2: Determine the Derivative of the Radical Part

Now we focus on the derivative of \(-\frac{x \sqrt{16-x^{2}}}{2}\). Given the product of two factors, we use the product rule which states \((fg)^{'} = f^{'}g+fg^{'}\). Therefore, it's necessary to compute the derivatives of \(f=-\frac{x}{2}\) and \(g=\sqrt{16-x^{2}}\) individually. Derivative of \(f\) is pretty straightforward: \(f^{'}=-\frac{1}{2}\). The derivative of \(g\) involves chain rule, we calculate it as \(g^{'} = \frac{-x}{\sqrt{16-x^{2}}}\). Now, using the product rule, we get \[-\frac{1}{2}\sqrt{16-x^{2}}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\]

Step 3: Combine the Partial Derivatives

Finally, combine the two partial derivatives previously calculated: \[y^{'}= \frac{2}{\sqrt{16-x^{2}}}-\frac{1}{2}\sqrt{16-x^{2}}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\]