Q3. Chain Rule of Derivatives 1-6 Write the composite function in the form \( f(g(x)) \). [Identify the inner function \( u=g(x) \) and the outer function \( y=f(u) \).\( ] Then find the derivative d y / d x \) \( \begin{array}{ll}\text { 1. } y=\sqrt[3]{1+4 x} & \text { 2. } y=\left(2 x^{3}+5\right)^{4} \\ \text { 3. } y=\tan \pi x & \text { 4. } y=\sin (\cot x) \\ \text { 5. } y=e^{\sqrt{x}} & \text { 6. } y=\sqrt{2-e^{x}}\end{array} \)

For \( y=\sqrt[3]{1+4x} \), you can express it as \( f(u) = u^{1/3} \) with \( g(x) = 1 + 4x \). The derivative \( \frac{dy}{dx} = \frac{1}{3}(1 + 4x)^{-2/3} \cdot 4 = \frac{4}{3(1 + 4x)^{2/3}} \). For \( y=(2x^{3}+5)^{4} \), let \( f(u) = u^{4} \) and \( g(x) = 2x^{3} + 5 \). Applying the chain rule, \( \frac{dy}{dx} = 4(2x^{3}+5)^{3} \cdot (6x^{2}) = 24x^{2}(2x^{3}+5)^{3} \). For the remaining functions, similar steps can be followed: 1. \( y = \tan(\pi x) \): \( f(u) = \tan(u) \), \( g(x) = \pi x \), \( \frac{dy}{dx} = \sec^2(\pi x) \cdot \pi \). 2. \( y = \sin(\cot x) \): \( f(u) = \sin(u) \), \( g(x) = \cot(x) \), \( \frac{dy}{dx} = \cos(\cot x)(-\csc^2 x) \). 3. \( y = e^{\sqrt{x}} \): \( f(u) = e^u \), \( g(x) = \sqrt{x} \), \( \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \). 4. \( y = \sqrt{2-e^{x}} \): \( f(u) = \sqrt{u} \), \( g(x) = 2-e^{x} \), \( \frac{dy}{dx} = \frac{1}{2\sqrt{2-e^{x}}}(-e^{x}) \). Now, go ahead and tackle more problems using this structure!