How do you implicitly differentiate $6 = y \frac{ln y}{x}$ $6=ylny/x$ ?

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Answer 1 · 2016-09-05T16:56:43

$\frac{d y}{d x} = \frac{6}{1 + ln y}, or, = 6 {(ln e y)}^{- 1}$ $dy/dx=6/(1+lny), or, =6(lney)^-1$ .

Rewriting the eqn. as, $: 6 x = y ln y$ $: 6x=ylny$

$:. ln(6x)=ln(ylny)$

$:. ln6+lnx=lny+ln(lny)$

$:. d/dx ln6+lnx=d/dx(lny+ln(lny))$

$:. 0+1/x=d/dy(lny+ln(lny))dy/dx........."[Chain Rule]"$

$:. 1/x={1/y+1/lnyd/dy(lny)}dy/dx$

$:. 1/x={(1/y+1/lny*1/y)}dy/dx$ .

$:. 1/x=1/y(1+1/lny)dy/dx=1/y((1+lny)/lny)dy/dx$

$=((1+lny)/(ylny))dy/dx$

$:. dy/dx=(ylny)/(x(1+lny))$ .

Remembering that, by the original eqn., $ylny=6x$ , we have,

$dy/dx=6/(1+lny), or, 6/(lne+lny)=6/ln(ey)=6(lney)^-1$ .

Alternatively,

$6x=ylny rArr d/dy(6x)=d/dy(ylny)$

$rArr 6dx/dy=yd/dy(lny)+(lny)*d/dy(y)$

$rArr 6dx/dy=y*1/y+lny=1+lny$

$rArr dx/dy=(1+lny)/6$

$rArr dy/dx=1/(dx/dy)=6/(1+lny)$ , as before.!

Enjoy maths.!

How do you implicitly differentiate 6=ylny/x6=ylnyx6=ylny/x?