1. Problem: Find the first derivatives of the given functions. 2. Formula: Use the product rule $\frac{d}{dx}[uv] = u'v + uv'$, chain rule $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$, and derivatives of basic functions: $\frac{d}{dx}[e^x] = e^x$, $\frac{d}{dx}[\sin x] = \cos x$, $\frac{d}{dx}[\cos x] = -\sin x$, $\frac{d}{dx}[\ln x] = \frac{1}{x}$, and $\frac{d}{dx}[x^n] = nx^{n-1}$. 3. (a) $f(x) = x^2 e^x$ \[f'(x) = \frac{d}{dx}[x^2] e^x + x^2 \frac{d}{dx}[e^x] = 2x e^x + x^2 e^x = e^x(2x + x^2)\] 4. (b) $f(x) = 2 \sin x \cos x$ \[f'(x) = 2(\cos x \cos x + \sin x (-\sin x)) = 2(\cos^2 x - \sin^2 x)\] 5. (c) $f(x) = \sin 2x$ \[f'(x) = \cos 2x \cdot \frac{d}{dx}[2x] = 2 \cos 2x\] 6. (d) $f(x) = x \sin ax$ \[f'(x) = 1 \cdot \sin ax + x \cdot a \cos ax = \sin ax + a x \cos ax\] 7. (e) $f(x) = e^{ax} \sin ax \tan^{-1} ax$ Using product rule for three functions: \[f'(x) = (e^{ax})' \sin ax \tan^{-1} ax + e^{ax} (\sin ax)' \tan^{-1} ax + e^{ax} \sin ax (\tan^{-1} ax)'\] Calculate each derivative: \[(e^{ax})' = a e^{ax}, \quad (\sin ax)' = a \cos ax, \quad (\tan^{-1} ax)' = \frac{a}{1 + a^2 x^2}\] So, \[f'(x) = a e^{ax} \sin ax \tan^{-1} ax + e^{ax} a \cos ax \tan^{-1} ax + e^{ax} \sin ax \frac{a}{1 + a^2 x^2}\] 8. (f) $f(x) = \ln(x^a + x^{-a})$ \[f'(x) = \frac{1}{x^a + x^{-a}} \cdot \frac{d}{dx}[x^a + x^{-a}] = \frac{1}{x^a + x^{-a}} (a x^{a-1} - a x^{-a-1})\] 9. (g) $f(x) = \ln(a x + a^{-x})$ \[f'(x) = \frac{1}{a x + a^{-x}} \cdot \frac{d}{dx}[a x + a^{-x}] = \frac{1}{a x + a^{-x}} (a - a^{-x} \ln a)\] 10. (h) $f(x) = x^x$ Rewrite as $f(x) = e^{x \ln x}$ \[f'(x) = e^{x \ln x} \cdot \frac{d}{dx}[x \ln x] = x^x (\ln x + 1)\] Final answers: (a) $f'(x) = e^x (2x + x^2)$ (b) $f'(x) = 2(\cos^2 x - \sin^2 x)$ (c) $f'(x) = 2 \cos 2x$ (d) $f'(x) = \sin ax + a x \cos ax$ (e) $f'(x) = a e^{ax} \sin ax \tan^{-1} ax + a e^{ax} \cos ax \tan^{-1} ax + \frac{a e^{ax} \sin ax}{1 + a^2 x^2}$ (f) $f'(x) = \frac{a x^{a-1} - a x^{-a-1}}{x^a + x^{-a}}$ (g) $f'(x) = \frac{a - a^{-x} \ln a}{a x + a^{-x}}$ (h) $f'(x) = x^x (\ln x + 1)$