Question

In Exercises \(33-38,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\cot \frac{\pi u}{10}, \quad u=g(x)=5 \sqrt{x}, \quad x=1$$

Short answer

The derivative of the composite function evaluated at \(x=1\) is computed using the chain rule, applied step-by-step according to the solution walkthrough. The exact value depends on computational accuracy, but the correct expression is \((f \circ g)'(1) = -\csc^2\left(\frac{\pi \cdot 5}{10}\right) \left(\frac{\pi}{10}\right) \cdot \left(\frac{5}{2}\right)\).

Step-by-step solution

Differentiating the functions

Firstly, differentiate function \(f(u)\) to find \(f'(u)\) and function \(g(x)\) to find \(g'(x)\). The derivative of the cotangent function is \(-\csc^2(x)\), while the derivative of the square root function is \(\frac{1}{2\sqrt{x}}\). So: \[f'(u)=-\csc^2\left(\frac{\pi u}{10}\right) \left(\frac{\pi}{10}\right) \, and \, \, g'(x)=\frac{5}{2 \sqrt{x}}\]

Apply the chain rule

According to the chain rule, \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x)\). We substitute \( f'(u) \) and \( g'(x) \) from step 1 into this formula, we get: \[ (f \circ g)'(x) = -\csc^2\left(\frac{\pi \cdot g(x)}{10}\right) \left(\frac{\pi}{10}\right) \cdot g'(x) \]

Substitute the given \(x\) value

Finally, substitute \(x = 1\) into the equation from step 2 to get the value of \((f \circ g)'(x)\) at \(x=1\): \[ (f \circ g)'(1) = -\csc^2\left(\frac{\pi \cdot g(1)}{10}\right) \left(\frac{\pi}{10}\right) \cdot g'(1) \]

Compute the value at \(x=1\)

Since \(g(x) = 5 \sqrt{x}\), \(g(1) = 5\). And since \(g'(x) = \frac{5}{2 \sqrt{x}}\), \(g'(1) = \frac{5}{2}\). Substitute these values into the equation from step 3 to compute the final value \[ (f \circ g)'(1) = -\csc^2\left(\frac{\pi \cdot 5}{10}\right) \left(\frac{\pi}{10}\right) \cdot \left(\frac{5}{2}\right) \]