Derivatives Exponential Logarithmic 289173

1. **Problem:** Find $\frac{dy}{dx}$ for $y = 3^{4x}$. **Step 1:** Use the exponential derivative formula $\frac{d}{dx} a^{u} = a^{u} \ln a \cdot \frac{du}{dx}$. $$\frac{dy}{dx} = 3^{4x} \ln 3 \cdot 4 = 4 \ln 3 \cdot 3^{4x}$$ 2. **Problem:** Find $\frac{dy}{dx}$ for $y = \frac{1 + 2^x}{1 - 2^x}$. **Step 1:** Use quotient rule: $\frac{d}{dx} \frac{f}{g} = \frac{f'g - fg'}{g^2}$. Let $f = 1 + 2^x$, $g = 1 - 2^x$. $$f' = 2^x \ln 2, \quad g' = -2^x \ln 2$$ $$\frac{dy}{dx} = \frac{(2^x \ln 2)(1 - 2^x) - (1 + 2^x)(-2^x \ln 2)}{(1 - 2^x)^2}$$ Simplify numerator: $$= \frac{2^x \ln 2 (1 - 2^x) + 2^x \ln 2 (1 + 2^x)}{(1 - 2^x)^2} = \frac{2^x \ln 2 [(1 - 2^x) + (1 + 2^x)]}{(1 - 2^x)^2}$$ $$= \frac{2^x \ln 2 (2)}{(1 - 2^x)^2} = \frac{2^{x+1} \ln 2}{(1 - 2^x)^2}$$ 3. **Problem:** Find $\frac{dy}{dx}$ for $y = 4^x \ln 4x$. **Step 1:** Use product rule: $\frac{d}{dx} (uv) = u'v + uv'$. Let $u = 4^x$, $v = \ln 4x$. $$u' = 4^x \ln 4, \quad v' = \frac{1}{4x} \cdot 4 = \frac{1}{x}$$ $$\frac{dy}{dx} = 4^x \ln 4 \cdot \ln 4x + 4^x \cdot \frac{1}{x} = 4^x \left(\ln 4 \ln 4x + \frac{1}{x}\right)$$ 4. **Problem:** Find $\frac{dy}{dx}$ for $y = e^{-4x}$. $$\frac{dy}{dx} = e^{-4x} \cdot (-4) = -4 e^{-4x}$$ 5. **Problem:** Find $\frac{dy}{dx}$ for $y = e^x e^{\ln x}$. Note $e^{\ln x} = x$, so $y = e^x x$. Use product rule: $$\frac{dy}{dx} = e^x x + e^x \cdot 1 = e^x (x + 1)$$ 6. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln \frac{e^{2x} - 1}{e^{2x} + 1}$. Use chain rule and quotient rule inside logarithm. $$y = \ln (u), \quad u = \frac{e^{2x} - 1}{e^{2x} + 1}$$ $$\frac{dy}{dx} = \frac{1}{u} \cdot u'$$ Compute $u'$: $$u' = \frac{(2 e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2 e^{2x})}{(e^{2x} + 1)^2}$$ Simplify numerator: $$= 2 e^{2x} (e^{2x} + 1) - 2 e^{2x} (e^{2x} - 1) = 2 e^{2x} [ (e^{2x} + 1) - (e^{2x} - 1)] = 2 e^{2x} (2) = 4 e^{2x}$$ So: $$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{4 e^{2x}}{(e^{2x} + 1)^2} = \frac{4 e^{2x}}{(e^{2x} + 1)^2} \cdot \frac{e^{2x} + 1}{e^{2x} - 1} = \frac{4 e^{2x} (e^{2x} + 1)}{(e^{2x} + 1)^2 (e^{2x} - 1)} = \frac{4 e^{2x}}{(e^{2x} + 1)(e^{2x} - 1)}$$ Note denominator is $e^{4x} - 1$, so: $$\frac{dy}{dx} = \frac{4 e^{2x}}{e^{4x} - 1}$$ 7. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln (e^x x^2)$. Use log properties: $$y = \ln e^x + \ln x^2 = x + 2 \ln x$$ $$\frac{dy}{dx} = 1 + \frac{2}{x}$$ 8. **Problem:** Find $\frac{dy}{dx}$ for $y = x^{2x}$. Rewrite: $$y = e^{2x \ln x}$$ $$\frac{dy}{dx} = y \cdot \frac{d}{dx} (2x \ln x)$$ Compute derivative inside exponent: $$\frac{d}{dx} (2x \ln x) = 2 \ln x + 2x \cdot \frac{1}{x} = 2 \ln x + 2$$ So: $$\frac{dy}{dx} = x^{2x} (2 \ln x + 2) = 2 x^{2x} (\ln x + 1)$$ 9. **Problem:** Find $\frac{dy}{dx}$ for $y = x^{e^x}$. Rewrite: $$y = e^{e^x \ln x}$$ $$\frac{dy}{dx} = y \cdot \frac{d}{dx} (e^x \ln x)$$ Use product rule inside exponent: $$\frac{d}{dx} (e^x \ln x) = e^x \ln x + e^x \cdot \frac{1}{x} = e^x \left(\ln x + \frac{1}{x}\right)$$ So: $$\frac{dy}{dx} = x^{e^x} e^x \left(\ln x + \frac{1}{x}\right)$$ 10. **Problem:** Find $\frac{dy}{dx}$ for $y = \log \sqrt[3]{\sqrt{12x}}$. Rewrite inside log: $$\sqrt[3]{\sqrt{12x}} = (12x)^{\frac{1}{6}}$$ So: $$y = \log (12x)^{\frac{1}{6}} = \frac{1}{6} \log (12x)$$ Derivative: $$\frac{dy}{dx} = \frac{1}{6} \cdot \frac{1}{12x \ln 10} \cdot 12 = \frac{1}{6x \ln 10}$$ 11. **Problem:** Find $\frac{dy}{dx}$ for $y = \log [(x-1)^3 (x + 2)^4]$. Use log properties: $$y = 3 \log (x-1) + 4 \log (x+2)$$ Derivative: $$\frac{dy}{dx} = \frac{3}{(x-1) \ln 10} + \frac{4}{(x+2) \ln 10} = \frac{3}{(x-1) \ln 10} + \frac{4}{(x+2) \ln 10}$$ 12. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln (x + 3)^4$. Use log power rule: $$y = 4 \ln (x + 3)$$ $$\frac{dy}{dx} = \frac{4}{x + 3}$$ 13. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln (x + \sqrt{x^2 + 1})$. Use chain rule: $$\frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} = \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} (x + \sqrt{x^2 + 1})}$$ Simplify numerator and denominator: $$= \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} (x + \sqrt{x^2 + 1})} = \frac{1}{\sqrt{x^2 + 1}}$$ 14. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln^4 (x + 3)$. Let $u = \ln (x + 3)$, so $y = u^4$. $$\frac{dy}{dx} = 4 u^3 \cdot \frac{1}{x + 3} = \frac{4 \ln^3 (x + 3)}{x + 3}$$ 15. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln \sqrt{\frac{a+bx}{a-bx}}$. Rewrite: $$y = \frac{1}{2} \ln \frac{a+bx}{a-bx}$$ $$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{\frac{d}{dx} \left(\frac{a+bx}{a-bx}\right)}{\frac{a+bx}{a-bx}}$$ Compute derivative inside: $$u = \frac{a+bx}{a-bx}, \quad u' = \frac{b(a-bx) - (a+bx)(-b)}{(a-bx)^2} = \frac{b(a-bx) + b(a+bx)}{(a-bx)^2} = \frac{2ab}{(a-bx)^2}$$ So: $$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{2ab/(a-bx)^2}{(a+bx)/(a-bx)} = \frac{1}{2} \cdot \frac{2ab}{(a-bx)^2} \cdot \frac{a-bx}{a+bx} = \frac{ab}{(a-bx)(a+bx)} = \frac{ab}{a^2 - b^2 x^2}$$ 16. **Problem:** Find $\frac{dy}{dx}$ for $y = \ln \left(\frac{1 - \sin x}{1 + \sin x}\right)^3$. Rewrite: $$y = 3 \ln \frac{1 - \sin x}{1 + \sin x} = 3 \left(\ln (1 - \sin x) - \ln (1 + \sin x)\right)$$ Derivative: $$\frac{dy}{dx} = 3 \left(\frac{-\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x}\right) = 3 \cos x \left(-\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}\right)$$ Combine terms: $$= 3 \cos x \left(-\frac{1 + \sin x + 1 - \sin x}{(1 - \sin x)(1 + \sin x)}\right) = -3 \cos x \frac{2}{1 - \sin^2 x} = -\frac{6 \cos x}{\cos^2 x} = -\frac{6}{\cos x} = -6 \sec x$$ **Final answers:** 1. $\frac{dy}{dx} = 4 \ln 3 \cdot 3^{4x}$ 2. $\frac{dy}{dx} = \frac{2^{x+1} \ln 2}{(1 - 2^x)^2}$ 3. $\frac{dy}{dx} = 4^x \left(\ln 4 \ln 4x + \frac{1}{x}\right)$ 4. $\frac{dy}{dx} = -4 e^{-4x}$ 5. $\frac{dy}{dx} = e^x (x + 1)$ 6. $\frac{dy}{dx} = \frac{4 e^{2x}}{e^{4x} - 1}$ 7. $\frac{dy}{dx} = 1 + \frac{2}{x}$ 8. $\frac{dy}{dx} = 2 x^{2x} (\ln x + 1)$ 9. $\frac{dy}{dx} = x^{e^x} e^x \left(\ln x + \frac{1}{x}\right)$ 10. $\frac{dy}{dx} = \frac{1}{6 x \ln 10}$ 11. $\frac{dy}{dx} = \frac{3}{(x-1) \ln 10} + \frac{4}{(x+2) \ln 10}$ 12. $\frac{dy}{dx} = \frac{4}{x + 3}$ 13. $\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + 1}}$ 14. $\frac{dy}{dx} = \frac{4 \ln^3 (x + 3)}{x + 3}$ 15. $\frac{dy}{dx} = \frac{ab}{a^2 - b^2 x^2}$ 16. $\frac{dy}{dx} = -6 \sec x$