Menu Close

Category: Differentiation

Question-193486

Question Number 193486 by Mingma last updated on 15/Jun/23 Answered by Subhi last updated on 15/Jun/23 $${x}^{{x}} .{ln}\left({x}\right)={ln}\left({y}\right) \\ $$$${ln}\left({x}^{{x}} .{ln}\left({x}\right)\right)={ln}\left({ln}\left({y}\right)\right)\Rrightarrow\:{xln}\left({x}\right)+{ln}\left({ln}\left({x}\right)\right)={ln}\left({ln}\left({y}\right)\right) \\ $$$${ln}\left({x}\right)+{x}.\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}.{ln}\left({x}\right)}=\frac{\mathrm{1}}{{y}.{ln}\left({y}\right)}.\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}}…

if-U-f-x-y-z-and-z-f-x-y-then-find-the-formula-d-2-u-dx-2-in-terms-of-derivetive-of-F-and-derivative-of-z-respectively-

Question Number 130976 by BHOOPENDRA last updated on 31/Jan/21 $${if}\:{U}={f}\left({x},{y},{z}\right){and}\:{z}={f}\left({x},{y}\right){then}\:{find}\: \\ $$$${the}\:{formula}\:\frac{{d}^{\mathrm{2}} {u}}{{dx}^{\mathrm{2}} }\:{in}\:{terms}\:{of}\:{derivetive} \\ $$$${of}\:{F}\:{and}\:{derivative}\:{of}\:{z}\:{respectively}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

nice-calculus-calculate-I-0-Si-x-x-3-2-dx-8pi-Si-x-0-x-sin-t-t-dt-

Question Number 130949 by mnjuly1970 last updated on 31/Jan/21 $$\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:{calculate}:\: \\ $$$$\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{S}{i}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{dx}\overset{???} {=}\sqrt{\mathrm{8}\pi}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{S}{i}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \frac{{sin}\left({t}\right)}{{t}}{dt} \\ $$ Answered…

Question-65367

Question Number 65367 by ajfour last updated on 29/Jul/19 Commented by ajfour last updated on 29/Jul/19 $$\mathrm{For}\:\mathrm{maximum}\:\mathrm{arc}\:\mathrm{length}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{origin}\:\mathrm{centered}\:\mathrm{circle}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{shown}\:\mathrm{unit}\:\mathrm{circle},\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$…