Recitation 12

Exercise: Find the derivative of the function. Simplify where possible.

1. $y=(\tan^{-1}x)^2$
2. $y=\sin^{-1}(2x+1)$
3. $y=x\sin^{-1}x+\sqrt{1-x^2}$
1. By the chain rule, $y'=2(\tan^{-1}x)\times\frac{1}{1+x^2}=\frac{2(\tan^{-1}x)}{1+x^2}$
2. By the chain rule, $y'=\frac{1}{\sqrt{1-(2x+1)^2}}\times 2=\frac{2}{\sqrt{1-(2x+1)^2}}$
3. By the product rule and the chain rule, $y'=\sin^{-1}x+x\frac{1}{\sqrt{1-x^2}}+\frac{1}{2}(1-x^2)^{-\frac{1}{2}}\times(-2x)=\sin^{-1}x$

Exercise: Find the limit.

1. $\lim_{x\to -1^+}\sin^{-1}x$
2. $\lim_{x\to\infty} arctan(e^x)$
1. $\lim_{x\to -1^+}\sin^{-1}x=arcsin(-1)=-\frac{\pi}{2}$
2. As $\lim_{x\to\infty} e^x=\infty$, $\lim_{x\to\infty} arctan(e^x)=\frac{\pi}{2}$.

Exercise: Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

1. $\lim_{x\to 0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}$
2. $\lim_{x\to 1^+}\ln x \tan(\pi x/2)$
1. Easy to observe that the limit has the form 0/0, by l’Hospital’s Rule, $\lim_{x\to 0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}=\lim_{x\to 0}\frac{-\sin x+x}{4x^3}$. Again, apply l’Hospital’s Rule for the same reason, $\lim_{x\to 0}\frac{-\sin x+x}{4x^3}=\lim_{x\to 0}\frac{-\cos x+1}{12x^2}$

. Once again, l’Hospital’s Rule, $\lim_{x\to 0}\frac{-\cos x+1}{12x^2}=\lim_{x\to 0}\frac{\sin x}{24x}=\frac{1}{24}$.

1. Since $\ln x \tan(\pi x/2)=\frac{\ln x}{\cot(\pi x/2)}$ has the form 0/0, we can apply l’Hospital’s Rule, $\lim_{x\to 1^+}\frac{\ln x}{\cot(\pi x/2)}=\lim_{x\to 1^+}\frac{1/x}{-\csc^2 (\pi x/2)\times \pi/2}=\lim_{x\to 1^+}-\frac{\sin^2(\pi x/2)}{\pi x/2}=-\frac{2}{\pi}$.