If xy =ex-y show thatdy/dx=(logx)/{log(xe)}2

• 82

I would say "this question is wonderful" for the ones who take 'log x as ln x' and write

d/dx(log x) = 1/x where it must be d/dx (ln x) = 1/x but only if in your question, one had to prove

dy/dx = (logx)(log e)/(log(xe))2

i hope you can understand what i meant to say :P

• -36

ylnx=x-y    ---1

lnxdy/dx+y/x=1-dy/dx

dy/dx(lnx+1)=1-x/y        ----3

we can see from 1 that

y=x/1+lnx

putting in 3

dy/dx=y-x/x

put the value of y

dy/dx=1/1+lnx

now 1-= lne

now u can see furthur to conversion

• -21

first take log on both sides

ylog(x)=(x-y)loge      since loge=1

ylogx=x-y

ylogx+y=x

y(logx+1)=x

y=x/(1+logx)      now solve by basic rule of diffrentiation ie. apply d/dx(u/v)

• 21
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