1. **Problem statement:** Find the derivative of the function $f$ for each given case. 2. **Recall the product rule:** For functions $u(x)$ and $v(x)$, the derivative of their product is $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$ 3. **a) For $f(x) = (3x + 1) \cdot e^x$:** - Let $u = 3x + 1$, so $u' = 3$. - Let $v = e^x$, so $v' = e^x$. Using the product rule: $$f'(x) = u'v + uv' = 3 e^x + (3x + 1) e^x = (3 + 3x + 1) e^x = (3x + 4) e^x.$$ 4. **b) For $f(x) = x \cdot \sqrt{x}$:** Rewrite $\sqrt{x} = x^{1/2}$, so $$f(x) = x \cdot x^{1/2} = x^{3/2}.$$ Derivative using power rule: $$f'(x) = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{1/2} = \frac{3}{2} \sqrt{x}.$$ 5. **c) For $f(x) = x^2 \cdot \sin(x)$:** - Let $u = x^2$, so $u' = 2x$. - Let $v = \sin(x)$, so $v' = \cos(x)$. Using the product rule: $$f'(x) = u'v + uv' = 2x \sin(x) + x^2 \cos(x).$$ 6. **d) For $f(x) = \frac{1}{\pi} \cdot \cos(\pi x)$:** - Constant factor $\frac{1}{\pi}$ remains. - Derivative of $\cos(\pi x)$ is $-\sin(\pi x) \cdot \pi$ by chain rule. So, $$f'(x) = \frac{1}{\pi} \cdot (-\sin(\pi x) \cdot \pi) = -\sin(\pi x).$$ 7. **e) For $f(x) = 2x \cdot e^{-3x}$:** - Let $u = 2x$, so $u' = 2$. - Let $v = e^{-3x}$, so $v' = e^{-3x} \cdot (-3) = -3 e^{-3x}$. Using the product rule: $$f'(x) = u'v + uv' = 2 e^{-3x} + 2x (-3 e^{-3x}) = 2 e^{-3x} - 6x e^{-3x} = (2 - 6x) e^{-3x}.$$ 8. **f) For $f(t) = 3 t^{-2} \cdot e^t$:** - Let $u = 3 t^{-2}$, so $u' = 3 \cdot (-2) t^{-3} = -6 t^{-3}$. - Let $v = e^t$, so $v' = e^t$. Using the product rule: $$f'(t) = u'v + uv' = (-6 t^{-3}) e^t + 3 t^{-2} e^t = (-6 t^{-3} + 3 t^{-2}) e^t.$$