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

a) $A=6f(x)-g(x)$ The derivative of $A$ is $A'=6f'(x)-g'(x)$ At $x=1$: $$A'(1)=6f'(1)-g'(1)=6\times\frac{1}{2}-(-4)=3+4=7$$ b) $A=f(x)g^2(x)$ The derivative of $A$ is $A'=f'(x)g^2(x)+f(x)\times2g(x)\times g'(x)$ At $x=0$: $$A'(0)=f'(0)g^2(0)+f(0)\times2g(0)\times g'(0)$$ $$=-3\times1^2+1\times2\times1\times\frac{1}{2}=-3+1=-2$$ c) $A=\frac{f(x)}{g(x)+1}$ The derivative of $A$ is $A'=\frac{f'(x)\Big(g(x)+1\Big)-f(x)\Big(g(x)+1\Big)'}{\Big(g(x)+1\Big)^2}=\frac{f'(x)\Big(g(x)+1\Big)-f(x)g'(x)}{\Big(g(x)+1\Big)^2}$ At $x=1$: $$A'(1)=\frac{f'(1)\Big(g(1)+1\Big)-f(1)g'(1)}{\Big(g(1)+1\Big)^2}=\frac{\frac{1}{2}(5+1)-3\times(-4)}{(5+1)^2}$$ $$=\frac{3+12}{36}=\frac{15}{36}=\frac{5}{12}$$ d) $A=f(g(x))$ The derivative of $A$ is $A'=f'(g(x))g'(x)$ At $x=0$: $$A'(0)=f'(g(0))g'(0)=f'(1)\times\frac{1}{2}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$$ e) $A=g(f(x))$ The derivative of $A$ is $A'=g'(f(x))f'(x)$ At $x=0$: $$A'(0)=g'(f(0))f'(0)=g'(1)\times(-3)=(-4)\times(-3)=12$$ f) $A=(x+f(x))^{3/2}$ The derivative of $A$ is $A'=\frac{3}{2}(x+f(x))^{1/2}(x+f(x))'=\frac{3}{2}(x+f(x))^{1/2}(1+f'(x))$ At $x=1$: $$A'(1)=\frac{3}{2}(1+f(1))^{1/2}(1+f'(1))=\frac{3}{2}(1+3)^{1/2}(1+\frac{1}{2})$$ $$A'(1)=\frac{3}{2}\times4^{1/2}\times\frac{3}{2}=\frac{9}{4}\times2=\frac{9}{2}$$ g) $A=f(x+g(x))$ The derivative of $A$ is $A'=f'(x+g(x))(x+g(x))'=f'(x+g(x))(1+g'(x))$ At $x=0$: $$A'(0)=f'(0+g(0))(1+g'(0))=f'(1)\times(1+\frac{1}{2})$$ $$A'(0)=\frac{1}{2}\times\frac{3}{2}=\frac{3}{4}$$