1. **Find the derivatives:** 1.1. Find $D_x (\ln 3 \sqrt[3]{x})$. Step 1: Rewrite the function using properties of logarithms: $$\ln 3 \sqrt[3]{x} = \ln 3 + \ln x^{1/3} = \ln 3 + \frac{1}{3} \ln x$$ Step 2: Differentiate term-by-term: $$\frac{d}{dx}(\ln 3) = 0$$ $$\frac{d}{dx}\left(\frac{1}{3} \ln x\right) = \frac{1}{3} \cdot \frac{1}{x} = \frac{1}{3x}$$ Answer: $$D_x (\ln 3 \sqrt[3]{x}) = \frac{1}{3x}$$ 1.2. Find $D_x \left[ \tan^{-1} e^x - \cos^{-1} \sqrt{1 - x^2} \right]$. Step 1: Differentiate $\tan^{-1} e^x$ using chain rule: $$\frac{d}{dx} \tan^{-1} e^x = \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{e^x}{1 + e^{2x}}$$ Step 2: Differentiate $- \cos^{-1} \sqrt{1 - x^2}$. Let $u = \sqrt{1 - x^2} = (1 - x^2)^{1/2}$. $$\frac{d}{dx} \cos^{-1} u = - \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$$ Calculate $\frac{du}{dx}$: $$\frac{du}{dx} = \frac{1}{2} (1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}}$$ Calculate $\sqrt{1 - u^2}$: $$1 - u^2 = 1 - (1 - x^2) = x^2$$ So, $$\frac{d}{dx} \cos^{-1} \sqrt{1 - x^2} = - \frac{1}{\sqrt{x^2}} \cdot \left(-\frac{x}{\sqrt{1 - x^2}}\right) = - \frac{1}{|x|} \cdot \left(-\frac{x}{\sqrt{1 - x^2}}\right) = \frac{1}{\sqrt{1 - x^2}}$$ Since $x \neq 0$ and considering domain, the derivative simplifies to: $$\frac{1}{\sqrt{1 - x^2}}$$ Step 3: Combine derivatives: $$D_x \left[ \tan^{-1} e^x - \cos^{-1} \sqrt{1 - x^2} \right] = \frac{e^x}{1 + e^{2x}} - \frac{1}{\sqrt{1 - x^2}}$$ --- 2. **Find $y'$ if $\tan(xy) = 3 \cos(e - \pi x) + \sqrt{x - y}$**. Step 1: Differentiate both sides with respect to $x$ using implicit differentiation. Left side: $$\frac{d}{dx} \tan(xy) = \sec^2(xy) \cdot \frac{d}{dx} (xy)$$ Use product rule on $xy$: $$\frac{d}{dx} (xy) = y + x \frac{dy}{dx} = y + x y'$$ So, $$\frac{d}{dx} \tan(xy) = \sec^2(xy) (y + x y')$$ Right side: $$\frac{d}{dx} \left[3 \cos(e - \pi x) + \sqrt{x - y}\right] = 3 \cdot (-\sin(e - \pi x)) \cdot (-\pi) + \frac{1}{2 \sqrt{x - y}} (1 - y')$$ Simplify: $$= 3 \pi \sin(e - \pi x) + \frac{1 - y'}{2 \sqrt{x - y}}$$ Step 2: Set derivatives equal: $$\sec^2(xy) (y + x y') = 3 \pi \sin(e - \pi x) + \frac{1 - y'}{2 \sqrt{x - y}}$$ Step 3: Group terms with $y'$ on one side: $$\sec^2(xy) x y' + \frac{y'}{2 \sqrt{x - y}} = 3 \pi \sin(e - \pi x) + \frac{1}{2 \sqrt{x - y}} - \sec^2(xy) y$$ Step 4: Factor out $y'$: $$y' \left( \sec^2(xy) x + \frac{1}{2 \sqrt{x - y}} \right) = 3 \pi \sin(e - \pi x) + \frac{1}{2 \sqrt{x - y}} - \sec^2(xy) y$$ Step 5: Solve for $y'$: $$y' = \frac{3 \pi \sin(e - \pi x) + \frac{1}{2 \sqrt{x - y}} - \sec^2(xy) y}{\sec^2(xy) x + \frac{1}{2 \sqrt{x - y}}}$$ --- 3. **Find $f^{(3)}(x)$ if $f(x) = e^{4x^2}$**. Step 1: Find first derivative: $$f'(x) = e^{4x^2} \cdot 8x$$ Step 2: Find second derivative using product rule: $$f''(x) = \frac{d}{dx} (8x e^{4x^2}) = 8 e^{4x^2} + 8x \cdot e^{4x^2} \cdot 8x = 8 e^{4x^2} + 64 x^2 e^{4x^2} = e^{4x^2} (8 + 64 x^2)$$ Step 3: Find third derivative: $$f^{(3)}(x) = \frac{d}{dx} \left(e^{4x^2} (8 + 64 x^2)\right)$$ Use product rule: $$= e^{4x^2} \cdot 8x \cdot 2 \cdot 4 (8 + 64 x^2) + e^{4x^2} \cdot (0 + 128 x)$$ More carefully: $$\frac{d}{dx} e^{4x^2} = e^{4x^2} \cdot 8x$$ So, $$f^{(3)}(x) = e^{4x^2} \cdot 8x (8 + 64 x^2) + e^{4x^2} \cdot 128 x = e^{4x^2} \left[8x (8 + 64 x^2) + 128 x\right]$$ Simplify inside brackets: $$8x \cdot 8 = 64 x$$ $$8x \cdot 64 x^2 = 512 x^3$$ Sum: $$64 x + 512 x^3 + 128 x = (64 x + 128 x) + 512 x^3 = 192 x + 512 x^3$$ Final answer: $$f^{(3)}(x) = e^{4x^2} (192 x + 512 x^3) = 64 x e^{4x^2} (3 + 8 x^2)$$ --- 4. **Evaluate limits:** 4.1. $$\lim_{x \to 0} \frac{e^x - x - 1}{x^2}$$ Step 1: Use Taylor expansion of $e^x$ near 0: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ Step 2: Substitute into numerator: $$e^x - x - 1 = \left(1 + x + \frac{x^2}{2} + \cdots\right) - x - 1 = \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ Step 3: Divide by $x^2$: $$\frac{e^x - x - 1}{x^2} = \frac{x^2/2 + x^3/6 + \cdots}{x^2} = \frac{1}{2} + \frac{x}{6} + \cdots$$ Step 4: Take limit as $x \to 0$: $$\lim_{x \to 0} \frac{e^x - x - 1}{x^2} = \frac{1}{2}$$ 4.2. $$\lim_{x \to +\infty} \left[ \cos \left(\frac{1}{x}\right) \right]^x$$ Step 1: Use the fact that $\cos(1/x) \to \cos(0) = 1$ as $x \to \infty$. Step 2: Rewrite limit as exponential: $$\left[ \cos \left(\frac{1}{x}\right) \right]^x = e^{x \ln \left(\cos \frac{1}{x}\right)}$$ Step 3: Use expansion for $\cos z$ near 0: $$\cos z = 1 - \frac{z^2}{2} + \cdots$$ So, $$\cos \frac{1}{x} = 1 - \frac{1}{2 x^2} + \cdots$$ Step 4: Use expansion for $\ln(1 + u)$ near 0: $$\ln(1 + u) \approx u - \frac{u^2}{2} + \cdots$$ Here, $$u = - \frac{1}{2 x^2} + \cdots$$ So, $$\ln \left(\cos \frac{1}{x}\right) \approx - \frac{1}{2 x^2}$$ Step 5: Multiply by $x$: $$x \cdot \ln \left(\cos \frac{1}{x}\right) \approx x \cdot \left(- \frac{1}{2 x^2}\right) = - \frac{1}{2 x}$$ Step 6: Take limit as $x \to \infty$: $$\lim_{x \to \infty} - \frac{1}{2 x} = 0$$ Step 7: Therefore, $$\lim_{x \to +\infty} \left[ \cos \left(\frac{1}{x}\right) \right]^x = e^0 = 1$$ --- 5. **Find the equation of the tangent line to $y = x^2 + 4x - 2$ where the normal line is parallel to $-x - 2y + 6 = 0$**. Step 1: Find slope of given line: Rewrite: $$-x - 2y + 6 = 0 \implies 2y = -x + 6 \implies y = -\frac{1}{2} x + 3$$ Slope is $m = -\frac{1}{2}$. Step 2: The normal line to the curve has slope $-\frac{1}{2}$. Step 3: Find derivative of curve (slope of tangent): $$y' = 2x + 4$$ Step 4: Slope of normal line is negative reciprocal of tangent slope: $$m_{normal} = -\frac{1}{m_{tangent}}$$ Given $m_{normal} = -\frac{1}{2}$, so $$-\frac{1}{m_{tangent}} = -\frac{1}{2} \implies m_{tangent} = 2$$ Step 5: Solve for $x$ where $y' = 2$: $$2x + 4 = 2 \implies 2x = -2 \implies x = -1$$ Step 6: Find $y$ at $x = -1$: $$y = (-1)^2 + 4(-1) - 2 = 1 - 4 - 2 = -5$$ Step 7: Equation of tangent line at $(-1, -5)$ with slope 2: $$y - (-5) = 2 (x - (-1))$$ $$y + 5 = 2 (x + 1)$$ $$y = 2x + 2 - 5 = 2x - 3$$ Answer: $$y = 2x - 3$$ --- 6. **Analyze $f(x) = -5x^3 - 15x^2$ for extrema, intervals of increase/decrease, concavity, and inflection points.** Step 1: Find first derivative: $$f'(x) = -15x^2 - 30x = -15x(x + 2)$$ Step 2: Find critical points by setting $f'(x) = 0$: $$-15x(x + 2) = 0 \implies x = 0 \text{ or } x = -2$$ Step 3: Determine intervals of increase/decrease by testing values: - For $x < -2$, pick $x = -3$: $$f'(-3) = -15(-3)(-3 + 2) = -15(-3)(-1) = -15 \cdot 3 = -45 < 0$$ (decreasing) - For $-2 < x < 0$, pick $x = -1$: $$f'(-1) = -15(-1)(-1 + 2) = -15(-1)(1) = 15 > 0$$ (increasing) - For $x > 0$, pick $x = 1$: $$f'(1) = -15(1)(1 + 2) = -15(1)(3) = -45 < 0$$ (decreasing) Step 4: Classify critical points: - At $x = -2$, $f'$ changes from negative to positive: local minimum. - At $x = 0$, $f'$ changes from positive to negative: local maximum. Step 5: Find second derivative: $$f''(x) = -30x - 30 = -30(x + 1)$$ Step 6: Find inflection points by setting $f''(x) = 0$: $$-30(x + 1) = 0 \implies x = -1$$ Step 7: Determine concavity intervals: - For $x < -1$, pick $x = -2$: $$f''(-2) = -30(-2 + 1) = -30(-1) = 30 > 0$$ (concave up) - For $x > -1$, pick $x = 0$: $$f''(0) = -30(0 + 1) = -30 < 0$$ (concave down) Step 8: Summary: - Relative minimum at $x = -2$. - Relative maximum at $x = 0$. - Increasing on $(-2, 0)$. - Decreasing on $(-\infty, -2)$ and $(0, \infty)$. - Concave up on $(-\infty, -1)$. - Concave down on $(-1, \infty)$. - Inflection point at $x = -1$. --- **Final answers:** 1. $D_x (\ln 3 \sqrt[3]{x}) = \frac{1}{3x}$ 2. $D_x \left[ \tan^{-1} e^x - \cos^{-1} \sqrt{1 - x^2} \right] = \frac{e^x}{1 + e^{2x}} - \frac{1}{\sqrt{1 - x^2}}$ 3. $$y' = \frac{3 \pi \sin(e - \pi x) + \frac{1}{2 \sqrt{x - y}} - \sec^2(xy) y}{\sec^2(xy) x + \frac{1}{2 \sqrt{x - y}}}$$ 4. $$f^{(3)}(x) = 64 x e^{4x^2} (3 + 8 x^2)$$ 5. $$\lim_{x \to 0} \frac{e^x - x - 1}{x^2} = \frac{1}{2}$$ 6. $$\lim_{x \to +\infty} \left[ \cos \left(\frac{1}{x}\right) \right]^x = 1$$ 7. Tangent line equation: $$y = 2x - 3$$ 8. For $f(x) = -5x^3 - 15x^2$: - Relative minimum at $x = -2$. - Relative maximum at $x = 0$. - Increasing on $(-2, 0)$. - Decreasing on $(-\infty, -2)$ and $(0, \infty)$. - Concave up on $(-\infty, -1)$. - Concave down on $(-1, \infty)$. - Inflection point at $x = -1$.