1. **State the problem:** Find the derivative of $y = 2x \sin x$ and evaluate it at $x = \frac{\pi}{2}$. 2. **Formula and rules:** Use the product rule for derivatives: If $y = u(x)v(x)$, then $$y' = u'v + uv'$$ 3. **Apply product rule:** Let $u = 2x$ and $v = \sin x$. Calculate derivatives: $$u' = 2$$ $$v' = \cos x$$ So, $$y' = 2 \sin x + 2x \cos x$$ 4. **Evaluate at $x = \frac{\pi}{2}$:** $$y'\left(\frac{\pi}{2}\right) = 2 \sin \frac{\pi}{2} + 2 \cdot \frac{\pi}{2} \cos \frac{\pi}{2}$$ Since $\sin \frac{\pi}{2} = 1$ and $\cos \frac{\pi}{2} = 0$, $$y'\left(\frac{\pi}{2}\right) = 2 \cdot 1 + \cancel{2 \cdot \frac{\pi}{2} \cdot 0} = 2$$ --- 1. **State the problem:** Find the gradient of the curve $y = 5e^{3x^3 + 2x}$ at $x=0$. 2. **Formula and rules:** Use the chain rule: If $y = 5e^{g(x)}$, then $$y' = 5e^{g(x)} \cdot g'(x)$$ 3. **Calculate $g'(x)$:** $$g(x) = 3x^3 + 2x$$ $$g'(x) = 9x^2 + 2$$ 4. **Find $y'$:** $$y' = 5e^{3x^3 + 2x} (9x^2 + 2)$$ 5. **Evaluate at $x=0$:** $$y'(0) = 5e^{0} (9 \cdot 0 + 2) = 5 \cdot 1 \cdot 2 = 10$$ **Final answers:** - Derivative of $y=2x \sin x$ at $x=\frac{\pi}{2}$ is $2$. - Gradient of $y=5e^{3x^3 + 2x}$ at $x=0$ is $10$.