1. Problem: Find the derivative $f'(x)$ for each given function without simplifying. 2. (a) Given $f(x) = 5x^{12} - \frac{3}{7}x^7 + x^6 - 10e^x + 5\sqrt[4]{x}$. Formula: Use power rule $\frac{d}{dx}x^n = nx^{n-1}$ and derivative of $e^x$ is $e^x$. Step: $$f'(x) = 5 \cdot 12 x^{11} - \frac{3}{7} \cdot 7 x^{6} + 6x^{5} - 10 e^x + 5 \cdot \frac{1}{4} x^{\frac{1}{4} - 1}$$ 3. (b) Given $f(x) = (x^4 - 3x^2 + 15)(x^2 + 6e^x)$. Formula: Product rule $\frac{d}{dx}[u v] = u'v + uv'$. Let $u = x^4 - 3x^2 + 15$, $v = x^2 + 6e^x$. Calculate derivatives: $$u' = 4x^3 - 6x$$ $$v' = 2x + 6 e^x$$ Apply product rule: $$f'(x) = (4x^3 - 6x)(x^2 + 6e^x) + (x^4 - 3x^2 + 15)(2x + 6 e^x)$$ 4. (c) Given $f(x) = \frac{x^2 + 2x + 7}{2x^3 - x}$. Formula: Quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$. Let $u = x^2 + 2x + 7$, $v = 2x^3 - x$. Calculate derivatives: $$u' = 2x + 2$$ $$v' = 6x^2 - 1$$ Apply quotient rule: $$f'(x) = \frac{(2x + 2)(2x^3 - x) - (x^2 + 2x + 7)(6x^2 - 1)}{(2x^3 - x)^2}$$ 5. (d) Given $f(x) = \ln \left( \frac{1 + \sin x}{1 - \sin x} \right)$. Formula: Derivative of $\ln g(x)$ is $\frac{g'(x)}{g(x)}$. Let $g(x) = \frac{1 + \sin x}{1 - \sin x}$. Calculate $g'(x)$ using quotient rule: $$g'(x) = \frac{(\cos x)(1 - \sin x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2}$$ Simplify numerator: $$\cos x (1 - \sin x) + \cos x (1 + \sin x) = \cos x (1 - \sin x + 1 + \sin x) = 2 \cos x$$ So, $$g'(x) = \frac{2 \cos x}{(1 - \sin x)^2}$$ Therefore, $$f'(x) = \frac{g'(x)}{g(x)} = \frac{2 \cos x}{(1 - \sin x)^2} \cdot \frac{1 - \sin x}{1 + \sin x} = \frac{2 \cos x}{(1 - \sin x)(1 + \sin x)}$$ 6. (e) Given $f(x) = (x^2 + 1) \tan^{-1}(3x) + x \sin^{-1}(2x)$. Formula: Use product rule and chain rule. Derivatives: $$\frac{d}{dx} \tan^{-1}(3x) = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}$$ $$\frac{d}{dx} \sin^{-1}(2x) = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}}$$ Apply product rule to first term: $$\frac{d}{dx}[(x^2 + 1) \tan^{-1}(3x)] = (2x) \tan^{-1}(3x) + (x^2 + 1) \frac{3}{1 + 9x^2}$$ Derivative of second term: $$\frac{d}{dx}[x \sin^{-1}(2x)] = 1 \cdot \sin^{-1}(2x) + x \cdot \frac{2}{\sqrt{1 - 4x^2}}$$ Combine: $$f'(x) = 2x \tan^{-1}(3x) + (x^2 + 1) \frac{3}{1 + 9x^2} + \sin^{-1}(2x) + \frac{2x}{\sqrt{1 - 4x^2}}$$