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How do you implicitly differentiate #6=ylny/x#?

Answer 1

Christopher Allers

#dy/dx=6/(1+lny), or, =6(lney)^-1#.

Rewriting the eqn. as, # : 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.!

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Answer 2

William Abdalla

To implicitly differentiate ( 6 = \frac{y \ln y}{x} ), follow these steps:

Begin by differentiating each term with respect to ( x ).
Apply the product rule when differentiating ( y \ln y ).
After differentiation, solve for ( \frac{dy}{dx} ).

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Answer from HIX Tutor

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