Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 209229 by Tawa11 last updated on 04/Jul/24

Commented by klipto last updated on 06/Jul/24

take in of both side  1. iny=e^x inx  ((d(iny))/dx)=v(du/dx)+u(dv/dx)  (1/y) (dy/dx)=e^x inx+(e^x /x)  (dy/dx)=(e^x inx+(e^x /x))×y  (dy/dx)=(e^x inx+(e^x /x))x^e^x    klipto−quanta

takeinofbothside1.iny=exinxd(iny)dx=vdudx+udvdx1ydydx=exinx+exxdydx=(exinx+exx)×ydydx=(exinx+exx)xexkliptoquanta

Commented by klipto last updated on 06/Jul/24

2.x^m y^n =(x+y)^(m+n)   take in of B.S  in(x^m y^n )=(m+n)in(x+y)  (quick one:y=in(x+y),?let u=x+y,(du/(dx ))=1+(dy/dx)  y=inu,(dy/du)=(1/u),(dy/dx)=(du/dx)×(dy/du)=[1+(dy/dx)](1/u)=[1+(dy/dx)](1/(x+y))    (m/x)+(n/y) (dy/dx)=(m+n)[1+(dy/dx)](1/(x+y))  (m/x)+(n/y) (dy/dx)=(m+n)[(1/(x+y))+(1/(x+y)) (dy/dx)]  (n/y) (dy/dx)−((m+n)/(x+y)) (dy/dx)=((m+n)/(x+y))−(m/x)  [(n/y)−((m+n)/(x+y))](dy/dx)=((m+n)/(x+y))−(m/x)  ((n(x+y)−y(m+n))/(y(x+y))) (dy/dx)=((x(m+n)−m(x+y))/(x(x+y)))  (dy/dx)=((x(m+n)−m(x+y))/(x(x+y)))÷((n(x+y)−y(m+n))/(y(x+y)))  (dy/dx)=((mx+nx−mx−my)/(x^2 +xy))×((xy+y^2 )/(nx+ny−my−ny))  (dy/dx)=((nx−my)/(x^2 +xy))×((xy+y^2 )/(nx−my))  (dy/dx)=((xy+y^2 )/(x^2 +xy))=((y(x+y))/(x(x+y)))=(y/x)  klipto−quanta

2.xmyn=(x+y)m+ntakeinofB.Sin(xmyn)=(m+n)in(x+y)(quickone:y=in(x+y),?letu=x+y,dudx=1+dydxy=inu,dydu=1u,dydx=dudx×dydu=[1+dydx]1u=[1+dydx]1x+ymx+nydydx=(m+n)[1+dydx]1x+ymx+nydydx=(m+n)[1x+y+1x+ydydx]nydydxm+nx+ydydx=m+nx+ymx[nym+nx+y]dydx=m+nx+ymxn(x+y)y(m+n)y(x+y)dydx=x(m+n)m(x+y)x(x+y)dydx=x(m+n)m(x+y)x(x+y)÷n(x+y)y(m+n)y(x+y)dydx=mx+nxmxmyx2+xy×xy+y2nx+nymynydydx=nxmyx2+xy×xy+y2nxmydydx=xy+y2x2+xy=y(x+y)x(x+y)=yxkliptoquanta

Answered by Spillover last updated on 05/Jul/24

  1)  y=x^e^x  Taking the natuaral logarithm of both sides we get    ln(y)=ln(x^e^x)    ln(y)=e^xln(x)    Differentiating both sides with respect to x will imply that    1/y(dy/dx)=e^x(1/x) + ln(xy)(e^x)    1/y(dy/dx)=e^x/x +ln(x)(e^x)    dy/dx=[e^x/x + ln(x)(e^x)](y)    But y=x^e^x    dy/dx=[e^x/x + ln(x)(e^x)](x^e^x)     *dy/dx=(x^e^x)(e^x)[1/x + ln(x)]*

1) y=x^e^x Taking the natuaral logarithm of both sides we get ln(y)=ln(x^e^x) ln(y)=e^xln(x) Differentiating both sides with respect to x will imply that 1/y(dy/dx)=e^x(1/x) + ln(xy)(e^x) 1/y(dy/dx)=e^x/x +ln(x)(e^x) dy/dx=[e^x/x + ln(x)(e^x)](y) But y=x^e^x dy/dx=[e^x/x + ln(x)(e^x)](x^e^x) *dy/dx=(x^e^x)(e^x)[1/x + ln(x)]*

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir, I appreciate.

Thankssir,Iappreciate.

Answered by Spillover last updated on 05/Jul/24

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir. I appreciate.

Thankssir.Iappreciate.

Answered by Spillover last updated on 05/Jul/24

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir.  I appreciate

Thankssir.Iappreciate

Answered by A5T last updated on 05/Jul/24

1.    y=x^e^x  ⇒log_x y=e^x ⇒((ln(y))/(ln(x)))=e^x   ⇒lny=e^x lnx  ⇒(1/y)×(dy/dx)=(e^x /x)+lnx×e^x   ⇒(dy/dx)=((x^e^x  ×e^x )/x)+x^e^x  ln(x)e^x =x^(e^x −1) e^x +x^e^x  ln(x)e^x

1.y=xexlogxy=exln(y)ln(x)=exlny=exlnx1y×dydx=exx+lnx×exdydx=xex×exx+xexln(x)ex=xex1ex+xexln(x)ex

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir. I appreciate

Thankssir.Iappreciate

Answered by A5T last updated on 05/Jul/24

2. ln(x^m y^n )=ln[(x+y)^(m+n) ]  ⇒mln(x)+nln(y)=(m+n)ln(x+y)  (m/x)+(n/y)×(dy/dx)=((m+n)/(x+y))+((m+n)/(x+y))×(dy/dx)  ⇒(dy/dx)((n/y)−((m+n)/(x+y)))=((m+n)/(x+y))−(m/x)=((nx−my)/(x(x+y)))  ⇒(dy/dx)(((nx−my)/(y(x+y))))=((nx−my)/(x(x+y)))⇒(dy/dx)=(y/x)

2.ln(xmyn)=ln[(x+y)m+n]mln(x)+nln(y)=(m+n)ln(x+y)mx+ny×dydx=m+nx+y+m+nx+y×dydxdydx(nym+nx+y)=m+nx+ymx=nxmyx(x+y)dydx(nxmyy(x+y))=nxmyx(x+y)dydx=yx

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir. I appreciate.

Thankssir.Iappreciate.

Answered by lepuissantcedricjunior last updated on 05/Jul/24

1)y=x^e^x  =>(dy/dx)=(e^(e^x lnx) )′=(e^x lnx)′y    ⇒(dy/dx)=(e^x lnx+(e^x /x))x^e^x    2)si x^m y^n =(x+y)^(m+n)   prouvons que (dy/dx)=(y/x)  x^m y^n =(x+y)^(m+n) =(x+y)^m (x+y)^n

1)y=xex=>dydx=(eexlnx)=(exlnx)ydydx=(exlnx+exx)xex2)sixmyn=(x+y)m+nprouvonsquedydx=yxxmyn=(x+y)m+n=(x+y)m(x+y)n

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir. I appreciate.

Thankssir.Iappreciate.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com