## University Calculus: Early Transcendentals (3rd Edition)

a) $1$ b) $6$ c) $1$ d) $-1/9$ e) $-40/3$ f) $-1/3$ g) $-4/9$
a) $$\Big(5f(x)-g(x)\Big)'=5f'(x)-g'(x)$$ Therefore, $$\Big(5f(1)-g(1)\Big)'=5f'(1)-g'(1)=5\times\frac{-1}{3}-\Big(-\frac{8}{3}\Big)=-\frac{5}{3}+\frac{8}{3}=1$$ b) $$\Big(f(x)g^3(x)\Big)'=f'(x)g^3(x)+f(x)\Big(g^3(x)\Big)'=f'(x)g^3(x)+3f(x)g^2(x)g'(x)$$ Therefore, $$\Big(f(0)g^3(0)\Big)'=f'(0)g^3(0)+3f(0)g^2(0)g'(0)=5\times1^3+3\times1\times1^2\times\frac{1}{3}$$ $$=5+1=6$$ c) $$\Big(\frac{f(x)}{g(x)+1}\Big)'=\frac{f'(x)\Big(g(x)+1\Big)-f(x)\Big(g(x)+1\Big)'}{\Big(g(x)+1\Big)^2}=\frac{f'(x)\Big(g(x)+1\Big)-f(x)g'(x)'}{\Big(g(x)+1\Big)^2}$$ Therefore, $$\Big(\frac{f(1)}{g(1)+1}\Big)'=\frac{f'(1)\Big(g(1)+1\Big)-f(1)g'(1)}{\Big(g(1)+1\Big)^2}$$ $$=\frac{-\frac{1}{3}(-4+1)-3\Big(-\frac{8}{3}\Big)}{(-4+1)^2}=\frac{1+8}{9}=1$$ d) $$\Big(f(g(x))\Big)'=f'(g(x))\times g'(x)$$ Therefore, $$\Big(f(g(0))\Big)'=f'(g(0))\times g'(0)=f'(1)\times\frac{1}{3}=\Big(-\frac{1}{3}\Big)\times\frac{1}{3}=-\frac{1}{9}$$ e) $$\Big(g(f(x))\Big)'=g'(f(x))\times f'(x)$$ Therefore, $$\Big(g(f(0))\Big)'=g'(f(0))\times f'(0)=g'(1)\times5=\Big(-\frac{8}{3}\Big)\times5=-\frac{40}{3}$$ f) $$\Bigg[\Big(x^{11}+f(x)\Big)^{-2}\Bigg]'=-2\Big(x^{11}+f(x)\Big)^{-3}\Big(x^{11}+f(x)\Big)'=-2\Big(x^{11}+f(x)\Big)^{-3}\Big(11x^{10}+f'(x)\Big)$$ Therefore, $$\Bigg[\Big(1^{11}+f(1)\Big)^{-2}\Bigg]'=-2\Big(1^{11}+f(1)\Big)^{-3}\Big(11\times1^{10}+f'(1)\Big)$$ $$=-2(1+3)^{-3}\Big(11-\frac{1}{3}\Big)=-2\times4^{-3}\Big(\frac{32}{3}\Big)=\frac{-64}{64\times3}=-\frac{1}{3}$$ g) $$\Big(f(x+g(x))\Big)'=f'(x+g(x))\times(x+g(x))'=f'(x+g(x))\times(1+g'(x))$$ Therefore, $$\Big(f(0+g(0))\Big)'=f'(0+g(0))\times(1+g'(0))=f'(1)\times\Big(1+\frac{1}{3}\Big)=-\frac{1}{3}\times\frac{4}{3}=-\frac{4}{9}$$