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

a) $A=\sqrt xf(x)$ The derivative of $A$ is $A'=\frac{1}{2\sqrt x}f(x)+\sqrt xf'(x)$ At $x=1$: $$A'(1)=\frac{1}{2\sqrt 1}f(1)+\sqrt 1f'(1)=\frac{1}{2}\times(-3)+1\times\frac{1}{5}$$ $$=-\frac{3}{2}+\frac{1}{5}=-\frac{13}{10}$$ b) $A=\sqrt{f(x)}$ The derivative of $A$ is $A'=\frac{1}{2\sqrt{f(x)}}f'(x)$ At $x=0$: $$A'(0)=\frac{1}{2\sqrt{f(0)}}f'(0)=\frac{1}{2\sqrt9}\times(-2)=-\frac{2}{2\times3}=-\frac{1}{3}$$ c) $A=f(\sqrt x)$ The derivative of $A$ is $A'=f'(\sqrt x)(\sqrt x)'=\frac{f'(\sqrt x)}{2\sqrt x}$ At $x=1$: $$A'(1)=\frac{f'(\sqrt 1)}{2\sqrt 1}=\frac{f'(1)}{2}=\frac{\frac{1}{5}}{2}=\frac{1}{10}$$ d) $A=f(1-5\tan x)$ The derivative of $A$ is $$A'=f'(1-5\tan x)\times(1-5\tan x)'$$ $$=f'(1-5\tan x)\times(0-5\sec^2x)=-5\sec^2xf'(1-5\tan x)$$ At $x=0$: $$A'(0)=-5\sec^20f'(1-5\tan 0)=-\frac{5}{\cos^20}\times f'(1-5\times0)$$ $$=-\frac{5}{1^2}\times f'(1)=-5\times\frac{1}{5}=-1$$ e) $A=\frac{f(x)}{2+\cos x}$ The derivative of $A$ is $$A'=\frac{f'(x)(2+\cos x)-f(x)(2+\cos x)'}{(2+\cos x)^2}$$ $$=\frac{f'(x)(2+\cos x)-f(x)(-\sin x)}{(2+\cos x)^2}$$ $$=\frac{f'(x)(2+\cos x)+f(x)\sin x}{(2+\cos x)^2}$$ At $x=0$: $$A'(0)=\frac{f'(0)(2+\cos 0)+f(0)\sin 0}{(2+\cos 0)^2}$$ $$=\frac{(-2)(2+1)+9\times0}{(2+1)^2}=\frac{-6}{9}=-\frac{2}{3}$$ f) $A=10\sin\Big(\frac{\pi x}{2}\Big)f^2(x)$ The derivative of $A$ is $$A'=10\Big[f^2(x)\cos\Big(\frac{\pi x}{2}\Big)\Big(\frac{\pi x}{2}\Big)'+\sin\Big(\frac{\pi x}{2}\Big)\times2f(x)f'(x)\Big]$$ $$A'=10\Big[f^2(x)\cos\Big(\frac{\pi x}{2}\Big)\frac{\pi}{2}+2f(x)f'(x)\sin\Big(\frac{\pi x}{2}\Big)\Big]$$ At $x=1$: $$A'(1)=10\Big[f^2(1)\cos\Big(\frac{\pi\times1}{2}\Big)\frac{\pi}{2}+2f(1)f'(1)\sin\Big(\frac{\pi \times1}{2}\Big)\Big]$$ $$A'(1)=10\Big[(-3)^2\cos\Big(\frac{\pi}{2}\Big)\frac{\pi}{2}+2\times(-3)\times(\frac{1}{5})\sin\Big(\frac{\pi}{2}\Big)\Big]$$ $$=10\Big[9\times0\times\frac{\pi}{2}-\frac{6}{5}\times1\Big]=10\Big[-\frac{6}{5}\Big]=-12$$