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 \( d y / d x \) \( \begin{array}{ll}\text { 1. } y=\sqrt[3]{1+4 x} & \text { 2. } y=\left(2 x^{3}+5\right)^{4} \\ \text { 3. } y=\tan \pi x & \text { 4. } y=\sin (\cot x) \\ \text { 5. } y=e^{\sqrt{x}} & \text { 6. } y=\sqrt{2-e^{x}}\end{array} \)

1. For \( y = \sqrt[3]{1 + 4x} \), let \( g(x) = 1 + 4x \) and \( f(u) = \sqrt[3]{u} \). Then, \( y = f(g(x)) = f(1 + 4x) \). To find the derivative, \( \frac{dy}{dx} = \frac{1}{3}(1 + 4x)^{-2/3} \cdot 4 = \frac{4}{3(1 + 4x)^{2/3}} \). 2. For \( y = (2x^3 + 5)^4 \), let \( g(x) = 2x^3 + 5 \) and \( f(u) = u^4 \). Thus, \( y = f(g(x)) = f(2x^3 + 5) \). The derivative is \( \frac{dy}{dx} = 4(2x^3 + 5)^3 \cdot (6x^2) = 24x^2(2x^3 + 5)^3 \). 3. For \( y = \tan(\pi x) \), let \( g(x) = \pi x \) and \( f(u) = \tan(u) \). Hence, \( y = f(g(x)) = f(\pi x) \). The derivative is \( \frac{dy}{dx} = \sec^2(\pi x) \cdot \pi = \pi \sec^2(\pi x) \). 4. For \( y = \sin(\cot x) \), let \( g(x) = \cot x \) and \( f(u) = \sin(u) \). Thus, \( y = f(g(x)) = f(\cot x) \). The derivative is \( \frac{dy}{dx} = \cos(\cot x) \cdot (-\csc^2(x)) = -\cos(\cot x) \csc^2(x) \). 5. For \( y = e^{\sqrt{x}} \), let \( g(x) = \sqrt{x} \) and \( f(u) = e^u \). Therefore, \( y = f(g(x)) = f(\sqrt{x}) \). The derivative is \( \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \). 6. For \( y = \sqrt{2 - e^x} \), let \( g(x) = 2 - e^x \) and \( f(u) = \sqrt{u} \). Thus, \( y = f(g(x)) = f(2 - e^x) \). The derivative is \( \frac{dy}{dx} = \frac{1}{2\sqrt{2 - e^x}} \cdot (-e^x) = -\frac{e^x}{2\sqrt{2 - e^x}} \).