1. Problem: Find the derivatives of the given functions. 2. Recall the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ and the derivative of $$x^x$$ using logarithmic differentiation. --- a) $$f(x) = x^6 + 2x - 1$$ Using the power rule: $$f'(x) = 6x^{5} + 2 - 0 = 6x^{5} + 2$$ --- b) $$h(x) = \frac{3}{5}x^{5} - 4x^{-2}$$ Derivative: $$h'(x) = \frac{3}{5} \cdot 5 x^{4} - 4 \cdot (-2) x^{-3} = 3x^{4} + 8x^{-3}$$ Intermediate step showing cancellation: $$h'(x) = \cancel{\frac{3}{5} \cdot 5} x^{4} + 8x^{-3} = 3x^{4} + 8x^{-3}$$ --- c) $$f(t) = \sqrt{t} + \frac{1}{t^{3}} = t^{\frac{1}{2}} + t^{-3}$$ Derivative: $$f'(t) = \frac{1}{2} t^{-\frac{1}{2}} - 3 t^{-4} = \frac{1}{2 \sqrt{t}} - 3 t^{-4}$$ --- d) $$f(x) = x^{x} + \frac{1}{x^{x}} = x^{x} + x^{-x}$$ Use logarithmic differentiation for $$x^{x}$$: Let $$y = x^{x}$$, then $$\ln y = x \ln x$$. Differentiate both sides: $$\frac{1}{y} \frac{dy}{dx} = \ln x + 1$$ So, $$\frac{dy}{dx} = y (\ln x + 1) = x^{x} (\ln x + 1)$$ Similarly for $$x^{-x}$$: Let $$z = x^{-x}$$, then $$\ln z = -x \ln x$$. Differentiate: $$\frac{1}{z} \frac{dz}{dx} = -\ln x - 1$$ So, $$\frac{dz}{dx} = z (-\ln x - 1) = x^{-x} (-\ln x - 1)$$ Therefore, $$f'(x) = x^{x} (\ln x + 1) + x^{-x} (-\ln x - 1)$$ --- Final answers: a) $$f'(x) = 6x^{5} + 2$$ b) $$h'(x) = 3x^{4} + 8x^{-3}$$ c) $$f'(t) = \frac{1}{2 \sqrt{t}} - 3 t^{-4}$$ d) $$f'(x) = x^{x} (\ln x + 1) + x^{-x} (-\ln x - 1)$$