Let's solve each problem step-by-step as requested: --- ### **8. \( f(x) = 5x^5 - 3x^4 + 10x/x^2 \) Find the derivative.** First, simplify the function: \[ f(x) = 5x^5 - 3x^4 + \frac{10x}{x^2} \] \[ = 5x^5 - 3x^4 + 10x^{-1} \] Now, differentiate term by term: - Derivative of \(5x^5\) is \(25x^4\) - Derivative of \(-3x^4\) is \(-12x^3\) - Derivative of \(10x^{-1}\) is \(-10x^{-2}\) So, \[ f'(x) = 25x^4 - 12x^3 - 10x^{-2} \] \[ \boxed{f'(x) = 25x^4 - 12x^3 - \frac{10}{x^2}} \] --- ### **9. \( f(x) = \sin(\cos(\sqrt{x})) \) Find the derivative.** Let \( u = \cos(\sqrt{x}) \), so \( f(x) = \sin(u) \). Chain rule: \[ f'(x) = \cos(u) \cdot u' \] \( u = \cos(\sqrt{x}) \), so \( u' = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \) Combine: \[ f'(x) = \cos(\cos(\sqrt{x})) \cdot \left[ -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \right] \] \[ \boxed{f'(x) = -\frac{\sin(\sqrt{x}) \cos(\cos(\sqrt{x}))}{2\sqrt{x}}} \] --- ### **10. \( f(x) = \cos x \cot x \) Find the derivative.** Product rule: If \( f(x) = u \cdot v \), then \( f'(x) = u'v + uv' \). Let \( u = \cos x \), \( v = \cot x \). - \( u' = -\sin x \) - \( v' = -\csc^2 x \) So, \[ f'(x) = (-\sin x)\cot x + \cos x(-\csc^2 x) \] \[ = -\sin x \cot x - \cos x \csc^2 x \] \[ \boxed{f'(x) = -\sin x \cot x - \cos x \csc^2 x} \] --- ### **11. \( f(x) = (7/x + x/7)^7 \) Find the derivative.** Let \( u = 7/x + x/7 \), so \( f(x) = u^7 \). Chain rule: \[ f'(x) = 7u^6 \cdot u' \] \( u = 7x^{-1} + \frac{1}{7}x \), so \( u' = -7x^{-2} + \frac{1}{7} \) So, \[ f'(x) = 7(7/x + x/7)^6 \left(-7x^{-2} + \frac{1}{7}\right) \] \[ \boxed{f'(x) = 7\left( \frac{7}{x} + \frac{x}{7} \right)^6 \left( -\frac{7}{x^2} + \frac{1}{7} \right)} \] --- ### **12. \( f(x) = qx + \sqrt{x} \) Find the derivative.** - Derivative of \(qx\) is \(q\) - Derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) \[ f'(x) = q + \frac{1}{2\sqrt{x}} \] \[ \boxed{f'(x) = q + \frac{1}{2\sqrt{x}}} \] --- ### **13. \( f(x) = x^3 - 1 \) at \( x = 2 \) Find the equation of the tangent line.** 1. \( f'(x) = 3x^2 \). At \( x = 2 \), \( f'(2) = 3(2^2) = 12 \) 2. \( f(2) = 2^3 - 1 = 7 \) 3. Point: \( (2, 7) \), Slope: 12 Equation: \[ y - 7 = 12(x - 2) \] \[ \boxed{y = 12x - 17} \] --- ### **14. \( y = \sin(2x) \) at \( x = 0 \) Find the equation of the tangent line.** 1. \( \frac{dy}{dx} = 2\cos(2x) \). At \( x = 0 \), \( 2\cos(0) = 2 \) 2. \( y(0) = \sin(0) = 0 \) 3. Point: \( (0, 0) \), Slope: 2 Equation: \[ y = 2x \] \[ \boxed{y = 2x} \] --- ### **15. \( x^2 + y^2 = 9 \) Find \( \frac{dy}{dx} \) using implicit differentiation.** Differentiate both sides with respect to \(x\): \[ 2x + 2y \frac{dy}{dx} = 0 \] \[ 2y \frac{dy}{dx} = -2x \] \[ \frac{dy}{dx} = -\frac{x}{y} \] \[ \boxed{\frac{dy}{dx} = -\frac{x}{y}} \] --- **Let me know if you'd like a more detailed explanation on any specific problem!**

8. \( f'(x) = 25x^4 - 12x^3 - \dfrac{10}{x^2} \) 9. \( f'(x) = -\dfrac{\sin(\sqrt{x}) \cos(\cos(\sqrt{x}))}{2\sqrt{x}} \) 10. \( f'(x) = -\sin x \cot x - \cos x \csc^2 x \) 11. \( f'(x) = 7 \left( \frac{7}{x} + \frac{x}{7} \right)^6 \left( -\frac{7}{x^2} + \frac{1}{7} \right) \) 12. \( f'(x) = q + \dfrac{1}{2\sqrt{x}} \) 13. \( y = 12x - 17 \) 14. \( y = 2x \) 15. \( \dfrac{dy}{dx} = -\dfrac{x}{y} \)