Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 48182 by cesar.marval.larez@gmail.com last updated on 20/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

$$1)t=\frac{x^m{y}^n}{{(x+y)}^{m+n}}$$$$lnt=mlnx+nlny-(m+n)ln(x+y)$$$$\frac{1}{t}\frac{dt}{dx}=\frac{m}{x}+\frac{n}{y}\times \frac{dy}{dx}-\frac{m+n}{x+y}(1+\frac{dy}{dx})$$$$\frac{1}{t}\times \frac{dt}{dx}=\frac{m}{x}-\frac{m+n}{x+y}+\frac{dy}{dx}(\frac{n}{y}-\frac{m+n}{x+y})$$$$\frac{dt}{dx}.=t[\frac{mx+my-mx-nx}{x(x+y)}+\frac{dy}{dx}\left(\frac{nx+ny-my-ny}{y(x+y}\right)]$$$$$$$$=\frac{x^m{y}^n}{{(x+y)}^{m+n}}[\frac{my-nx}{x(x+y)}-\frac{dy}{dx}\left(\frac{my-nx}{y(x+y)}\right]$$$$=\frac{x^m{y}^n}{{(x+y)}^{m+n}}\times \frac{my-nx}{x+y}[\frac{1}{x}-\frac{dy}{dx}\times \frac{1}{y}]$$$$2)t=\sqrt{\frac{{(3x+1)}^4\times {(4x+1)}^5}{{(x^2+x+1)}^3}}$$$$lnt=\frac{1}{2}\times [4ln(3x+1)+5ln(4x+1)-3ln(x^2+x+1)]$$$$\frac{1}{t}\times \frac{dt}{dx}=\frac{1}{2}[\frac{4}{3x+1}\times 3+\frac{5}{4x+1}\times 4-\frac{3}{x^2+x+1}\times (2x+1)]$$$$\frac{dt}{dx}=\sqrt{\frac{{(3x+1)}^4\times {(4x+1)}^5}{{(x^2+x+1)}^3}}\times \frac{1}{2}[\frac{12}{3x+1}+\frac{20}{4x+1}-\frac{3(2x+1)}{{(x^2+x+1)}^{}}]$$

Commented by cesar.marval.larez@gmail.com last updated on 21/Nov/18

$$Thankuverymuchiamlearning$$$$more.Ifidontknowsomeipost$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com