Calculus Basic Problems Part 2

Published 11/21/2025

\text{Find }y'\text{ using logarithmic differentiation if } y=\dfrac{(x^{2}+1)^{5/2}}{x(1-x^{2})}.\\[6pt] \textbf{Answer: }y'=\dfrac{(x^{2}+1)^{5/2}}{x(1-x^{2})}\left(\dfrac{5x}{x^{2}+1}-\dfrac{1}{x}+\dfrac{2x}{1-x^{2}}\right).

\text{Find }y'\text{ using logarithmic differentiation if } y=\dfrac{(1+x)^{3}(1-x)^{2}}{\sqrt{1+x^{2}}}.\\[6pt] \textbf{Answer: }y'=\dfrac{(1+x)^{3}(1-x)^{2}}{\sqrt{1+x^{2}}}\left(\dfrac{3}{1+x}-\dfrac{2}{1-x}-\dfrac{x}{1+x^{2}}\right).

\text{Find }y'\text{ using logarithmic differentiation if } y=\dfrac{x^{4}\sqrt{1-x^{2}}}{(2+x^{3})^{2}}.\\[6pt] \textbf{Answer: }y'=\dfrac{x^{4}\sqrt{1-x^{2}}}{(2+x^{3})^{2}}\left(\dfrac{4}{x}-\dfrac{x}{1-x^{2}}-\dfrac{6x^{2}}{2+x^{3}}\right).

\text{Find }y'\text{ using logarithmic differentiation if } y=\dfrac{\sqrt[3]{x^{2}-1}}{x^{2}(1+x^{2})^{3/2}}.\\[6pt] \textbf{Answer: }y'=\dfrac{\sqrt[3]{x^{2}-1}}{x^{2}(1+x^{2})^{3/2}}\left(\dfrac{2x}{3(x^{2}-1)}-\dfrac{2}{x}-\dfrac{3x}{1+x^{2}}\right).

\text{Let }x_{1},\dots,x_{n}>0.\ \text{Use Jensen's inequality for the concave function }\ln x\text{ to show that}\\[4pt] \frac{1}{n}\sum_{k=1}^{n}\ln x_{k}\le \ln\!\Bigl(\frac{1}{n}\sum_{k=1}^{n}x_{k}\Bigr).\\[6pt] \textbf{Answer: }\displaystyle \frac{1}{n}\sum_{k=1}^{n}\ln x_{k}\le \ln\!\Bigl(\frac{1}{n}\sum_{k=1}^{n}x_{k}\Bigr), \ \text{with equality iff }x_{1}=\dots=x_{n}.

\text{Let }x_{1},\dots,x_{n}>0\text{ with }\sum_{k=1}^{n}x_{k}=1.\ \text{Use Jensen's inequality for the convex function }\\[2pt] f(x)=\dfrac{1}{1+x}\text{ on }(0,\infty)\text{ to obtain a lower bound for }\sum_{k=1}^{n}\frac{1}{1+x_{k}}.\\[6pt] \textbf{Answer: }\displaystyle \sum_{k=1}^{n}\frac{1}{1+x_{k}}\ge \frac{n^{2}}{n+1}, \ \text{with equality iff }x_{1}=\dots=x_{n}=\frac{1}{n}.

\text{Let }x_{1},\dots,x_{n}\in[0,\pi]\text{ and set }\bar x=\dfrac{1}{n}\sum_{k=1}^{n}x_{k}.\ \text{Using that }\sin x\text{ is concave on }[0,\pi],\\[2pt] \text{apply Jensen's inequality to compare }\dfrac{1}{n}\sum_{k=1}^{n}\sin x_{k}\text{ and }\sin\bar x.\\[6pt] \textbf{Answer: }\displaystyle \frac{1}{n}\sum_{k=1}^{n}\sin x_{k}\le \sin\Bigl(\frac{1}{n}\sum_{k=1}^{n}x_{k}\Bigr), \ \text{with equality iff }x_{1}=\dots=x_{n}\text{ or all }x_{k}\in\{0,\pi\}.

\text{Let }N=\sqrt{5}+\sqrt{6}+\cdots+\sqrt{13}.\ \text{Show that the integer part of }N\text{ lies between }26\text{ and }27.\\[4pt] \textbf{Outline: }\\ \text{(1) Compare each term }\sqrt{k}\text{ with suitable constants to get an upper bound, e.g. } \sqrt{k}<\sqrt{13}\text{ for all }k\le13,\\ \text{or group terms and compare with simple numbers like }3,4,5\text{ to show }N<27.\\ \text{(2) Similarly, use }\sqrt{k}>\sqrt{4}=2\text{ for }k\ge5\text{ or sharper estimates (e.g. }\sqrt{9}=3\text{) to obtain }N>26.\\ \text{(3) Conclude }26<N<27,\text{ so the integer part of }N\text{ is }26.\\[4pt] \textbf{Answer: }\lfloor N\rfloor=26.