Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 5672 by sanusihammed last updated on 23/May/16

$$Differentiate\frac{lnx}{e^x}fomthefirstprinciple.$$$$$$$$Pleasehelpme.$$

Answered by Yozzii last updated on 23/May/16

$$Lety=\frac{lnx}{e^x}.Fromfirstprinciples$$$$\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$$$$=\underset{h\to 0}{\lim }\frac{e^{-x-h}ln(x+h)-e^{-x}lnx}{h}$$$$=\underset{h\to 0}{\lim }(\frac{ln(x+h)}{{he}^{x+h}}-\frac{lnx}{{he}^x})$$$$=\underset{h\to 0}{\lim }\frac{ln(x+h)-e^{h}lnx}{{he}^x{e}^h}$$$$=e^{-x}\underset{h\to 0}{\lim }\frac{1}{{he}^h}(ln(x+h)-{lnx}^{e^h})$$$$=e^{-x}\underset{h\to 0}{\lim }\frac{1}{{he}^h}ln\frac{x+h}{x^{e^h}}$$$$\frac{dy}{dx}=e^{-x}\underset{h\to 0}{\lim }ln{(x^{1-e^h}+\frac{h}{x^{e^h}})}^{1/{he}^h}$$$$ln\frac{dy}{dx}=ln\left\{e^{-x}\right\}+ln\left(\underset{h\to 0}{\lim }\left\{\frac{1}{e^h}ln\left[{(x^{1-e^h}+\frac{h}{x^{e^h}})}^{1/h}\right]\right\}\right)$$$$ln\frac{dy}{dx}=-x+ln\left(\underset{h\to 0}{\lim }\frac{1}{h}{lnx}^{1-e^h}(1+\frac{h}{x})\right)$$$$ln\frac{dy}{dx}=-x+ln\left(\underset{h\to 0}{\lim }(\frac{1-e^h}{h}lnx+\frac{1}{h}ln(1+\frac{h}{x}))\right)$$$$$$$$\bullet eisanumbersuchthatforubeing$$$$anynumber,e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+...+\frac{u^n}{n!}+...$$$$\bullet Forj\in (-1,1]onecanwrite$$$$ln(1+j)=j-\frac{j^2}{2}+\frac{j^3}{3}-\frac{j^4}{4}+...+\frac{{(-1)}^{n+1}{j}^n}{n}+...$$$$$$$$\therefore ln\frac{dy}{dx}=-x+ln\{\left(lnx\right)\underset{h\to 0}{\lim }\frac{1}{h}(1-1-h-\frac{h^2}{2!}-\frac{h^3}{3!}-...)+\underset{h\to 0}{\lim }\frac{1}{h}(\frac{h}{x}-\frac{h^2}{2x^2}+\frac{h^3}{3x^3}-...)\}$$$$ln\frac{dy}{dx}=-x+ln\{\left(lnx\right)\underset{h\to 0}{\lim }(-1-\frac{h}{2!}-\frac{h^2}{3!}-...)+\underset{h\to 0}{\lim }(\frac{1}{x}-\frac{h}{2x^2}+\frac{h^2}{3x^3}-...)\}$$$$ln\frac{dy}{dx}=-x+ln\{\left(lnx\right)(-1)+\frac{1}{x}\}$$$$\frac{dy}{dx}=exp(-x+ln(x^{-1}-lnx))=e^{-x}\times (x^{-1}-lnx)$$$$$$$$$$$$$$$$$$

Commented by sanusihammed last updated on 24/May/16

$$Thisisgreat.wow.$$

Commented by sanusihammed last updated on 24/May/16

$$Thankssomuch.$$

Commented by Yozzii last updated on 23/May/16

$$Alternatively,onecouldprovethe$$$$quotientrulefromfirstprinciples$$$$andsubsequentlyapplyit.$$$$$$$$Letr\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}whereuandvarefunctions$$$$ofxandv\left(x\right)\ne 0.$$$$\therefore r(x+h)=\frac{u(x+h)}{v(x+h)}(h=\delta x)$$$$\therefore r(x+h)-r\left(x\right)=\frac{u(x+h)}{v(x+h)}-\frac{u\left(x\right)}{v\left(x\right)}=\delta r\left(x\right)$$$$=\frac{v\left(x\right)(u(x+h)-u\left(x\right)+u\left(x\right))-u\left(x\right)(v(x+h)-v\left(x\right)+v\left(x\right))}{v(x+h)v\left(x\right)}$$$$\delta r\left(x\right)=\frac{v\left(x\right)\delta u\left(x\right)+v\left(x\right)u\left(x\right)-u\left(x\right)\delta v\left(x\right)-u\left(x\right)v\left(x\right)}{v(x+h)v\left(x\right)}$$$$\delta r\left(x\right)=\frac{v\left(x\right)\delta u\left(x\right)-u\left(x\right)\delta v\left(x\right)}{v(x+\delta x)v\left(x\right)}$$$$Bydefinitionofthederivative,$$$$\frac{dr}{dx}=\underset{\delta x\to 0}{\lim }\frac{\delta r\left(x\right)}{\delta x}=\underset{\delta x\to 0}{\lim }\frac{1}{\delta x}\left(\frac{v\left(x\right)\delta u\left(x\right)-u\left(x\right)\delta v\left(x\right)}{v(x+\delta x)v\left(x\right)}\right)$$$$\frac{dr}{dx}=\underset{\delta x\to 0}{\lim }\frac{v\left(x\right)\frac{\delta u\left(x\right)}{\delta x}-u\left(x\right)\frac{\delta v\left(x\right)}{\delta x}}{v(x+\delta x)v\left(x\right)}$$$$Sincev\left(x\right)\ne 0forallx\in \mathbb{R},bythealgebra$$$$oflimits,$$$$\Rightarrow \frac{dr}{dx}=\frac{\underset{\delta x\to 0}{\lim }(v\left(x\right)\frac{\delta u\left(x\right)}{\delta x}-u\left(x\right)\frac{\delta v\left(x\right)}{\delta x})}{\underset{\delta x\to 0}{\lim }\left\{v(x+\delta x)v\left(x\right)\right\}}$$$$\frac{dr}{dx}=\frac{v\left(x\right)\left(\underset{\delta x\to 0}{\lim }\frac{\delta u\left(x\right)}{\delta x}\right)-u\left(x\right)\left(\underset{\delta x\to 0}{\lim }\frac{\delta v\left(x\right)}{\delta x}\right)}{\underset{\delta x\to 0}{\lim }\left\{v(x+\delta x)v\left(x\right)\right\}}$$$$Now,wesubstitutethefollowingresults:$$$$\underset{\delta x\to 0}{\lim }\frac{\delta u\left(x\right)}{\delta x}=\frac{du}{dx},\underset{\delta x\to 0}{\lim }\frac{\delta v\left(x\right)}{\delta x}=\frac{dv}{dx}\mathrm{\&}$$$$\underset{\delta x\to 0}{\lim }\left\{v(x+\delta x)v\left(x\right)\right\}=v(x+0)v\left(x\right)=v^2\left(x\right).$$$$\therefore \frac{dr}{dx}=\frac{v\left(x\right)\frac{du}{dx}-u\left(x\right)\frac{dv}{dx}}{v^2\left(x\right)}◻$$$$$$$$So,ifr\left(x\right)=\frac{lnx}{e^x}$$$$\Rightarrow \frac{dr}{dx}=\frac{e^x{x}^{-1}-e^{x}lnx}{e^{2x}}=\frac{x^{-1}-lnx}{e^x}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com