1. A a) Given $f(x)=\ln(2x^3)$. Use the chain rule with $u=2x^3$ so $u'=6x^2$. Then $f'(x)=\frac{u'}{u}=\frac{6x^2}{2x^3}=\frac{3}{x}$. 2. A b) Given $f(x)=e^{x^7}$. Differentiate the exponential: $f'(x)=e^{x^7}\cdot (x^7)'$. Thus $f'(x)=e^{x^7}\cdot 7x^6=7x^6e^{x^7}$. 3. A c) Given $f(x)=\ln(11x^7)$. Let $u=11x^7$ so $u'=77x^6$. Then $f'(x)=\frac{u'}{u}=\frac{77x^6}{11x^7}=\frac{7}{x}$. 4. A d) Given $f(x)=e^{x^2}+x^3$. Differentiate termwise: $(e^{x^2})'=e^{x^2}\cdot 2x$ and $(x^3)'=3x^2$. Therefore $f'(x)=2xe^{x^2}+3x^2$. 5. A e) Given $f(x)=\log_e(7x^{-2})$. Rewrite as $f(x)=\ln(7x^{-2})$ and let $u=7x^{-2}$ so $u'=-14x^{-3}$. Then $f'(x)=\frac{u'}{u}=\frac{-14x^{-3}}{7x^{-2}}=-\frac{2}{x}$. 6. A f) Given $f(x)=e^{-x}$. Differentiate the exponential: $f'(x)=e^{-x}\cdot(-1)$. So $f'(x)=-e^{-x}$. 7. A g) Given $f(x)=\ln(e^x+x^3)$. Use the chain rule with $u=e^x+x^3$ so $u'=e^x+3x^2$. Then $f'(x)=\frac{u'}{u}=\frac{e^x+3x^2}{e^x+x^3}$. 8. A h) Given $f(x)=\ln(e^x x^3)$. Use log rules to simplify: $\ln(e^x x^3)=\ln(e^x)+\ln(x^3)=x+3\ln x$. Differentiate to get $f'(x)=1+\frac{3}{x}$. 9. A i) Given $f(x)=\ln\left(\frac{x^2+1}{x^3-x}\right)$. Use log properties: $f(x)=\ln(x^2+1)-\ln(x^3-x)$ and differentiate termwise. Thus $f'(x)=\frac{2x}{x^2+1}-\frac{3x^2-1}{x^3-x}$. 10. B 1) Given $y=4^{4^{x^4}}$. Set $u=4^{x^4}$ so $y=4^u$ and compute $u'=4^{x^4}\ln 4\cdot 4x^3$. Then $\frac{y'}{y}=u'\ln 4$ so $y'=4^{4^{x^4}}\cdot 4^{x^4}\cdot 4x^3\cdot (\ln 4)^2$. 11. B 2) Given $y=4^{-5x^3}$. Use $y= e^{\ln 4\cdot(-5x^3)}$ or direct rule $y'=y\ln 4\cdot(-15x^2)$. So $y'=4^{-5x^3}\cdot (-15x^2)\ln 4$. 12. B 3) Given $y=\log_3(3x^2)$. Write $y=\dfrac{\ln(3x^2)}{\ln 3}$ and differentiate: $(\ln(3x^2))'=\dfrac{6x}{3x^2}=\dfrac{2}{x}$. Hence $y'=\dfrac{2}{x\ln 3}$. 13. B 4) Given $y=\log_2(4x^2)$. Write $y=\dfrac{\ln(4x^2)}{\ln 2}$ and compute $(\ln(4x^2))'=\dfrac{8x}{4x^2}=\dfrac{2}{x}$. Thus $y'=\dfrac{2}{x\ln 2}$. 14. B 5) Given $y=\log_3\bigl(3x^5+5\bigr)^5$. Use $y=\dfrac{\ln((3x^5+5)^5)}{\ln 3}=\dfrac{5\ln(3x^5+5)}{\ln 3}$ and differentiate: $(\ln(3x^5+5))'=\dfrac{15x^4}{3x^5+5}$. So $y'=\dfrac{5\cdot 15x^4}{(3x^5+5)\ln 3}=\dfrac{75x^4}{(3x^5+5)\ln 3}$. 15. B 6) Given $y=\log_5(-5x^3-2)^3$. Write $y=\dfrac{\ln((-5x^3-2)^3)}{\ln 5}=\dfrac{3\ln(-5x^3-2)}{\ln 5}$ and differentiate: $(\ln(-5x^3-2))'=\dfrac{-15x^2}{-5x^3-2}=\dfrac{15x^2}{5x^3+2}$. Therefore $y'=\dfrac{3\cdot 15x^2}{(5x^3+2)\ln 5}=\dfrac{45x^2}{(5x^3+2)\ln 5}$. 16. B 7) Given $y=(4^{x^3}+2)^3$. Use chain rule: $y'=3(4^{x^3}+2)^2\cdot (4^{x^3})'$, and $(4^{x^3})'=4^{x^3}\ln 4\cdot 3x^2$. Thus $y'=9x^2\ln 4\cdot 4^{x^3}(4^{x^3}+2)^2$. 17. B 8) Given $y=3^{(x^4+1)^3}$. Differentiate: $y'=3^{(x^4+1)^3}\ln 3\cdot 3(x^4+1)^2\cdot 4x^3$. So $y'=12x^3(x^4+1)^2\ln 3\cdot 3^{(x^4+1)^3}$. 18. B 9) Given $y=3^{\cos(3x^4)}$. Use $y'=3^{\cos(3x^4)}\ln 3\cdot (\cos(3x^4))'$ and $(\cos(3x^4))'=-\sin(3x^4)\cdot 12x^3$. Hence $y'=-12x^3\ln 3\cdot \sin(3x^4)\cdot 3^{\cos(3x^4)}$. 19. B 10) Given $y=\log_5(\tan(4x^4))$. Write $y=\dfrac{\ln(\tan(4x^4))}{\ln 5}$ and differentiate: $(\ln(\tan(4x^4)))'=\dfrac{\sec^2(4x^4)\cdot 16x^3}{\tan(4x^4)}$. So $y'=\dfrac{16x^3\sec^2(4x^4)}{\tan(4x^4)\ln 5}$. 20. C 1) Given $f(x)=\tan x\sec^2 x$. Use product rule with $u=\tan x$ and $v=\sec^2 x$, where $u'=\sec^2 x$ and $v'=2\sec^2 x\tan x$. Thus $f'(x)=\sec^2 x\cdot\sec^2 x+\tan x\cdot 2\sec^2 x\tan x=\sec^4 x+2\tan^2 x\sec^2 x$. 21. C 2) Given $f(x)=\sec^3 x\sin^2 x$. Product rule with $(\sec^3 x)'=3\sec^3 x\tan x$ and $(\sin^2 x)'=2\sin x\cos x$. So $f'(x)=3\sec^3 x\tan x\sin^2 x+2\sec^3 x\sin x\cos x$. 22. C 3) Given $f(x)=\tan^5(x^3+x)$. Use chain rule: outer derivative $5\tan^4(u)\sec^2(u)$ with $u=x^3+x$ and $u'=3x^2+1$. Therefore $f'(x)=5\tan^4(x^3+x)\sec^2(x^3+x)\cdot(3x^2+1)$. 23. C 4) Given $y=\tan^{-1}(3x-5)$. Derivative of arctan is $\dfrac{1}{1+t^2}$ times inner derivative, so $y'=\dfrac{3}{1+(3x-5)^2}$. 24. C 5) Given $g(x)=\tan^{-1}(x^2)$. Differentiate: $g'(x)=\dfrac{1}{1+(x^2)^2}\cdot 2x=\dfrac{2x}{1+x^4}$. 25. C 6) Given $f(x)=\tan^3 x\cos^2 4x$. Product rule with $(\tan^3 x)'=3\tan^2 x\sec^2 x$ and $(\cos^2 4x)'=2\cos 4x\cdot(-\sin 4x)\cdot 4=-8\cos 4x\sin 4x$. So $f'(x)=3\tan^2 x\sec^2 x\cos^2 4x-8\tan^3 x\cos 4x\sin 4x$. 26. C 7) Given $f(x)=\sin^4\sqrt{x}$. Let $u=\sqrt{x}=x^{1/2}$ so $u'=\dfrac{1}{2\sqrt{x}}$, and outer derivative $4\sin^3 u\cos u$. Hence $f'(x)=4\sin^3(\sqrt{x})\cos(\sqrt{x})\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{2\sin^3(\sqrt{x})\cos(\sqrt{x})}{\sqrt{x}}$. 27. C 8) Given $f(x)=\sin^{-1}\sqrt{x}$. With $u=\sqrt{x}$ and $u'=\dfrac{1}{2\sqrt{x}}$ the derivative is $f'(x)=\dfrac{1}{\sqrt{1-u^2}}\cdot u'$. So $f'(x)=\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$. 28. C 9) Given $f(x)=\sin^{-1}\left(\dfrac{1}{3}x\right)$. Differentiate: $f'(x)=\dfrac{1}{\sqrt{1-(x/3)^2}}\cdot \dfrac{1}{3}=\dfrac{1}{3\sqrt{1-x^2/9}}$, which simplifies to $\dfrac{1}{\sqrt{9-x^2}}$ if desired. 29. C 10) Given $y=x^2\arctan(x^2)$. Use product rule: $y'=2x\arctan(x^2)+x^2\cdot \dfrac{1}{1+x^4}\cdot 2x$. Thus $y'=2x\arctan(x^2)+\dfrac{2x^3}{1+x^4}$.