2. If $p=\frac{q^{2}}{\pi}, q=3 \cos (r)+\sin (r), r=\ln \left(s^{\pi}\right)$ find $\frac{d p}{d s}$ at $s=e$. (Answer must be exact. 6 marks)

Step 1 ：First, substitute the expression for r into the expression for q: \(q = 3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi))\)

Step 2 ：Now, substitute the expression for q into the expression for p: \(p = \frac{(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi)))^2}{\pi}\)

Step 3 ：To find \(\frac{dp}{ds}\), we need to differentiate p with respect to s: \(\frac{dp}{ds} = \frac{d}{ds} \left(\frac{(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi)))^2}{\pi}\right)\)

Step 4 ：Apply the chain rule: \(\frac{dp}{ds} = \frac{2(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi)))}{\pi} \cdot \frac{d}{ds}(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi)))\)

Step 5 ：Differentiate the expression inside the parentheses with respect to s: \(\frac{d}{ds}(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi))) = -3 \sin (\ln (s^\pi)) \cdot \frac{\pi}{s} + \cos (\ln (s^\pi)) \cdot \frac{\pi}{s}\)

Step 6 ：Substitute this expression back into the expression for \(\frac{dp}{ds}\): \(\frac{dp}{ds} = \frac{2(3 \cos (\ln (s^\pi)) + \sin (\ln (s^\pi)))}{\pi} \cdot \left(-3 \sin (\ln (s^\pi)) \cdot \frac{\pi}{s} + \cos (\ln (s^\pi)) \cdot \frac{\pi}{s}\right)\)

Step 7 ：Now, substitute s = e into the expression for \(\frac{dp}{ds}\): \(\frac{dp}{ds} = \frac{2(3 \cos (\ln (e^\pi)) + \sin (\ln (e^\pi)))}{\pi} \cdot \left(-3 \sin (\ln (e^\pi)) \cdot \frac{\pi}{e} + \cos (\ln (e^\pi)) \cdot \frac{\pi}{e}\right)\)

Step 8 ：Simplify the expression: \(\frac{dp}{ds} = \frac{2(3 \cos (\pi) + \sin (\pi))}{\pi} \cdot \left(-3 \sin (\pi) \cdot \frac{\pi}{e} + \cos (\pi) \cdot \frac{\pi}{e}\right)\)

Step 9 ：Evaluate the trigonometric functions: \(\frac{dp}{ds} = \frac{2(-3 + 0)}{\pi} \cdot \left(0 - \frac{\pi}{e}\right)\)

Step 10 ：Finally, simplify the expression: \(\frac{dp}{ds} = \boxed{\frac{6\pi}{e}}\)