1. **Find the exact coordinates of the stationary point on the curve** $y = xe^x$. 2. **Find the coordinates of the points on the curve** $y = \frac{(x - 1)^2}{2x + 5}$ **where the tangent is parallel to the x-axis**. 3. **For the curve** $y = x^2 \ln 5x$, **find** $\frac{dy}{dx}$ **and** $\frac{d^2y}{dx^2}$ **at** $x=2$. 4. **Find the gradient of the curve** $y = 3 \sin 2x - 5 \tan x$ **at** $x=0$. 5. **Given parametric equations** $x = 3(1 + \sin^2 t)$, $y = 2 \cos^3 t$, **find** $\frac{dy}{dx}$ **in terms of** $t$. 6. **Find the equation of the tangent to the curve** $y = \tan^{-1} \left(\frac{x}{2}\right)$ **at** $x=2$. 7. **Differentiate with respect to** $x$: $x \tan^{-1} x$. --- ### Step 1: Stationary point of $y = xe^x$ - Stationary points occur where $\frac{dy}{dx} = 0$. - Differentiate using product rule: $\frac{dy}{dx} = e^x + xe^x = e^x(1 + x)$. - Set derivative to zero: $e^x(1 + x) = 0$. - Since $e^x \neq 0$, solve $1 + x = 0 \Rightarrow x = -1$. - Find $y$ at $x = -1$: $y = (-1)e^{-1} = -\frac{1}{e}$. - **Stationary point:** $\boxed{\left(-1, -\frac{1}{e}\right)}$. ### Step 2: Tangent parallel to x-axis for $y = \frac{(x - 1)^2}{2x + 5}$ - Tangent parallel to x-axis means $\frac{dy}{dx} = 0$. - Use quotient rule: $\frac{dy}{dx} = \frac{(2(x-1))(2x+5) - (x-1)^2(2)}{(2x+5)^2}$. - Simplify numerator: $$2(x-1)(2x+5) - 2(x-1)^2 = 2(x-1)[(2x+5) - (x-1)] = 2(x-1)(x+6)$$ - Set numerator zero: $2(x-1)(x+6) = 0 \Rightarrow x=1$ or $x=-6$. - Find $y$ at $x=1$: $y = \frac{0}{7} = 0$. - Find $y$ at $x=-6$: $y = \frac{(-7)^2}{2(-6)+5} = \frac{49}{-7} = -7$. - **Points:** $\boxed{(1,0)}$ and $\boxed{(-6,-7)}$. ### Step 3: Derivatives of $y = x^2 \ln 5x$ at $x=2$ - Use product rule: $\frac{dy}{dx} = 2x \ln 5x + x^2 \cdot \frac{1}{5x} \cdot 5 = 2x \ln 5x + x$. - Simplify: $\frac{dy}{dx} = 2x \ln 5x + x$. - Differentiate again for second derivative: $$\frac{d^2y}{dx^2} = 2 \ln 5x + 2x \cdot \frac{1}{x} + 1 = 2 \ln 5x + 2 + 1 = 2 \ln 5x + 3$$ - Evaluate at $x=2$: $$\frac{dy}{dx} = 2(2) \ln 10 + 2 = 4 \ln 10 + 2$$ $$\frac{d^2y}{dx^2} = 2 \ln 10 + 3$$ - **Answers:** $\frac{dy}{dx} = 4 \ln 10 + 2$, $\frac{d^2y}{dx^2} = 2 \ln 10 + 3$. ### Step 4: Gradient of $y = 3 \sin 2x - 5 \tan x$ at $x=0$ - Differentiate: $$\frac{dy}{dx} = 3 \cdot 2 \cos 2x - 5 \sec^2 x = 6 \cos 2x - 5 \sec^2 x$$ - Evaluate at $x=0$: $$6 \cos 0 - 5 \sec^2 0 = 6(1) - 5(1) = 1$$ - **Gradient:** $\boxed{1}$. ### Step 5: Parametric derivative for $x=3(1+\sin^2 t)$, $y=2 \cos^3 t$ - Compute $\frac{dx}{dt} = 3 \cdot 2 \sin t \cos t = 6 \sin t \cos t$. - Compute $\frac{dy}{dt} = 2 \cdot 3 \cos^2 t (-\sin t) = -6 \cos^2 t \sin t$. - Then: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-6 \cos^2 t \sin t}{6 \sin t \cos t}$$ - Cancel common factors: $$\frac{\cancel{-6} \cos^2 t \cancel{\sin t}}{\cancel{6} \cancel{\sin t} \cos t} = - \frac{\cos^2 t}{\cos t} = - \cos t$$ - **Simplified:** $\boxed{-\cos t}$. ### Step 6: Tangent to $y = \tan^{-1} \left(\frac{x}{2}\right)$ at $x=2$ - Differentiate: $$\frac{dy}{dx} = \frac{1}{1 + (x/2)^2} \cdot \frac{1}{2} = \frac{1}{2(1 + x^2/4)} = \frac{1}{2 + \frac{x^2}{2}} = \frac{2}{4 + x^2}$$ - Evaluate at $x=2$: $$\frac{dy}{dx} = \frac{2}{4 + 4} = \frac{2}{8} = \frac{1}{4}$$ - Find $y$ at $x=2$: $$y = \tan^{-1} (1) = \frac{\pi}{4}$$ - Equation of tangent line: $$y - \frac{\pi}{4} = \frac{1}{4}(x - 2)$$ - **Answer:** $\boxed{y = \frac{1}{4}x + \frac{\pi}{4} - \frac{1}{2}}$. ### Step 7: Differentiate $x \tan^{-1} x$ - Use product rule: $$\frac{d}{dx} (x \tan^{-1} x) = \tan^{-1} x + x \cdot \frac{1}{1 + x^2} = \tan^{-1} x + \frac{x}{1 + x^2}$$ --- **Summary:** - Stationary point: $\left(-1, -\frac{1}{e}\right)$ - Tangent horizontal points: $(1,0)$ and $(-6,-7)$ - Derivatives at $x=2$: $\frac{dy}{dx} = 4 \ln 10 + 2$, $\frac{d^2y}{dx^2} = 2 \ln 10 + 3$ - Gradient at $x=0$: $1$ - Parametric derivative: $-\cos t$ - Tangent line at $x=2$: $y = \frac{1}{4}x + \frac{\pi}{4} - \frac{1}{2}$ - Derivative of $x \tan^{-1} x$: $\tan^{-1} x + \frac{x}{1 + x^2}$