1. Problem: Find the implicit derivatives for the given equations. (c) $$\sqrt{x} + y = x^4 + y^4$$ 2. Formula and rules: For implicit differentiation, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (i.e., use chain rule for $y$ terms: $\frac{d}{dx}[y] = y'$). 3. Differentiate each term: - $$\frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}}$$ - $$\frac{d}{dx}[y] = y'$$ - $$\frac{d}{dx}[x^4] = 4x^3$$ - $$\frac{d}{dx}[y^4] = 4y^3 y'$$ 4. Write the differentiated equation: $$\frac{1}{2\sqrt{x}} + y' = 4x^3 + 4y^3 y'$$ 5. Group $y'$ terms on one side: $$y' - 4y^3 y' = 4x^3 - \frac{1}{2\sqrt{x}}$$ 6. Factor out $y'$: $$y'(1 - 4y^3) = 4x^3 - \frac{1}{2\sqrt{x}}$$ 7. Solve for $y'$: $$y' = \frac{4x^3 - \frac{1}{2\sqrt{x}}}{1 - 4y^3}$$ --- 1. Problem: Find the tangent line equation at point $(2,1)$ for the curve (c) $$x^2 + 2xy + 4y^2 = 12$$ (ellipse) 2. Differentiate implicitly with respect to $x$: $$2x + 2(y + x y') + 8y y' = 0$$ 3. Expand: $$2x + 2y + 2x y' + 8y y' = 0$$ 4. Group $y'$ terms: $$2x y' + 8y y' = -2x - 2y$$ 5. Factor $y'$: $$y'(2x + 8y) = -2x - 2y$$ 6. Solve for $y'$: $$y' = \frac{-2x - 2y}{2x + 8y} = \frac{-(x + y)}{x + 4y}$$ 7. Evaluate at $(2,1)$: $$y' = \frac{-(2 + 1)}{2 + 4} = \frac{-3}{6} = -\frac{1}{2}$$ 8. Equation of tangent line: $$y - y_0 = m(x - x_0)$$ $$y - 1 = -\frac{1}{2}(x - 2)$$ --- 1. Problem: Find the second derivative $y''$ for (c) $$\sin y + \cos x = 1$$ 2. Differentiate implicitly once: $$\cos y \cdot y' - \sin x = 0$$ 3. Solve for $y'$: $$y' = \frac{\sin x}{\cos y}$$ 4. Differentiate $y'$ to find $y''$: $$y'' = \frac{d}{dx} \left( \frac{\sin x}{\cos y} \right)$$ 5. Use quotient rule or product rule: $$y'' = \frac{\cos x \cdot \cos y - \sin x (-\sin y) y'}{\cos^2 y}$$ 6. Substitute $y'$: $$y'' = \frac{\cos x \cos y + \sin x \sin y \cdot \frac{\sin x}{\cos y}}{\cos^2 y} = \frac{\cos x \cos y + \sin^2 x \tan y}{\cos^2 y}$$ 7. Simplify: $$y'' = \frac{\cos x \cos y}{\cos^2 y} + \frac{\sin^2 x \tan y}{\cos^2 y} = \frac{\cos x}{\cos y} + \frac{\sin^2 x \sin y}{\cos^3 y}$$