Solution:

Let \(f(x) = 4x - \cos(x)\) and \(g(x) = 9x + 2\). Then:

\[f'(x) = 4 + \sin(x)\] \[g'(x) = 9\]

Using the quotient rule:

\[\begin{align}y' &= \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2}\\ &= \frac{(4 + \sin(x))(9x + 2) - 9(4x - \cos(x))}{(9x + 2)^2} \end{align}\]

Expanding the numerator:

\[y'= \frac{36x + 8 + 9x\sin(x) + 2\sin(x) - 36x + 9\cos(x)}{(9x + 2)^2}\]

Simplifying:

\[y'= \frac{8 + \sin(x)(9x + 2) + 9\cos(x)}{(9x + 2)^2}\]

Solution:

First, rewrite using exponent notation: \(y = \frac{ax^{1/3}}{a^2e^x + ax}\)

Let \(f(x) = ax^{1/3}\) and \(g(x) = a^2e^x + ax\). Then:

\[f'(x) = \frac{a}{3}x^{-2/3} = \frac{a}{3x^{2/3}}\] \[g'(x) = a^2e^x + a\]

Using the quotient rule:

\[y' = \frac{\frac{a}{3x^{2/3}}(a^2e^x + ax) - (a^2e^x + a)(ax^{1/3})}{(a^2e^x + ax)^2}\]

Factor and simplify the numerator:

\[\begin{align}y'&= \frac{\frac{a^3e^x}{3x^{2/3}} + \frac{a^2x}{3x^{2/3}} - a^3e^x x^{1/3} - a^2x^{1/3}}{(a^2e^x + ax)^2}\\ &= \frac{\frac{a^3e^x}{3\sqrt[3]{x^2}} + \frac{a^2\sqrt[3]{x}}{3} - a^3e^x\sqrt[3]{x} - a^2\sqrt[3]{x}}{(a^2e^x + ax)^2}\end{align}\]

Combining like terms and factoring:

\[\begin{align} y' &= \frac{a^3e^x\left(\frac{1}{3\sqrt[3]{x^2}} - \sqrt[3]{x}\right) + a^2\sqrt[3]{x}\left(\frac{1}{3} - 1\right)}{(a^2e^x + ax)^2}\\ &= \frac{a^3e^x\left(\frac{1 - 3x}{3\sqrt[3]{x^2}}\right) - \frac{2a^2\sqrt[3]{x}}{3}}{(a^2e^x + ax)^2}\end{align}\]

This can be simplified further to:

\[y' = \frac{ae^x(1 - 3x) - 2x}{3\sqrt[3]{x^2}(ae^x + x)^2}\]

Solution:

Let \(f(x) = 11\sin(x)\) and \(g(x) = \tan(x)\). Then:

\[f'(x) = 11\cos(x)\] \[g'(x) = \sec^2(x)\]

Using the product rule:

\[y' = 11\cos(x)\tan(x) + 11\sec^2(x)\sin(x)\]

Evaluate at \(x = \frac{\pi}{3}\):

\[y'\left(\frac{\pi}{3}\right) = 11\cos\left(\frac{\pi}{3}\right)\tan\left(\frac{\pi}{3}\right) + 11\sec^2\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3}\right)\]

Using known values: \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\), \(\sec\left(\frac{\pi}{3}\right) = 2\), \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\):

\[\begin{align}y'\left(\frac{\pi}{3}\right)&= 11 \cdot \frac{1}{2} \cdot \sqrt{3} + 11 \cdot 2^2 \cdot \frac{\sqrt{3}}{2}\\ &= \frac{11\sqrt{3}}{2} + \frac{44\sqrt{3}}{2}\\ &= \frac{55\sqrt{3}}{2}\end{align}\]

Solution:

Let \(f(x) = 6x^8\) and \(g(x) = \sec(x)\). Then:

\[f'(x) = 48x^7\] \[g'(x) = \sec(x)\tan(x)\]

Using the product rule:

\[y' = 48x^7\sec(x) + 6x^8\sec(x)\tan(x)\]

Evaluate the slope at \(x = \pi\). Note that \(\sec(\pi) = \frac{1}{\cos(\pi)} = \frac{1}{-1} = -1\) and \(\tan(\pi) = 0\):

\[y'(\pi) = 48\pi^7\cdot (-1) + 6\pi^8\cdot (-1)\cdot 0 = -48\pi^7\]

Find the \(y\)-coordinate at \(x = \pi\):

\[y(\pi) = 6\pi^8\sec(\pi) = 6\pi^8\cdot(-1) = -6\pi^8\]

So the point is \((\pi, -6\pi^8)\).

Using the point-slope form \(y - y_1 = m(x - x_1)\):

\[y - (-6\pi^8) = -48\pi^7(x - \pi)\] \[\begin{align}&\implies y + 6\pi^8 = -48\pi^7x + 48\pi^8\\ &\implies y = -48\pi^7x + 48\pi^8 - 6\pi^8\\ &\implies y = -48\pi^7x + 42\pi^8\end{align}\]

Solution:

Let \(f(x) = 4\cot(x)\) and \(g(x) = 5 + 8\cos(x)\). Then:

\[f'(x) = -4\csc^2(x)\] \[g'(x) = -8\sin(x)\]

Using the quotient rule:

\[y' = \frac{-4\csc^2(x)(5 + 8\cos(x)) - (-8\sin(x))(4\cot(x))}{(5 + 8\cos(x))^2}\]

Evaluate at \(x = \frac{\pi}{2}\). Note that \(\cot\left(\frac{\pi}{2}\right) = 0\), \(\csc\left(\frac{\pi}{2}\right) = 1\), \(\cos\left(\frac{\pi}{2}\right) = 0\), and \(\sin\left(\frac{\pi}{2}\right) = 1\). So,:

\[\begin{align}y'\left(\frac{\pi}{2}\right) &= \frac{-4\cdot 1^2\cdot (5 + 8 \cdot 0) - (-8 \cdot 1)\cdot(4 \cdot 0)}{(5 + 8 \cdot 0)^2}\\ &= \frac{-4\cdot (5 + 0) - 0}{5^2}\\ &= \frac{-20}{25}\\ &= -\frac{4}{5}\end{align}\]