How do you find the implicit differentiation of #1+lnxy=3^(x-y)#?

2 Answers

Given:
#1+\ln(xy)=3^{x-y}\ .......(1)#

#1+\ln(x)+\ln(y)=3^{x-y}#

differentiating w.r.t. #x# as follows

#\frac{d}{dx}(1+\ln(x)+\ln(y))=\frac{d}{dx}(3^{x-y})#

#0+\frac1x+\frac1y\frac{dy}{dx}=3^{x-y}\ln(3)\frac{d}{dx}(x-y)#

#\frac1x+\frac1y\frac{dy}{dx}=3^{x-y}\ln(3)(1-\frac{dy}{dx})#

#\frac1x+\frac1y\frac{dy}{dx}=(1+\ln(xy))\ln(3)(1-\frac{dy}{dx})# (from(1))

#[(1+\ln(xy))\ln(3)+\frac1y]\frac{dy}{dx}=(1+\ln(xy))\ln(3)-\frac1x#

#\frac{dy}{dx}=\frac{(1+\ln(xy))\ln(3)-\frac1x}{(1+\ln(xy))\ln(3)+\frac1y}#

Jun 24, 2018

#dy/dx=((ln(3)*3^(x+y))/(3^(2y))-1/x)/(((ln(3)*3^(x+y))/3^(2y)+1/y)#

Explanation:

First separate the equation into easier pieces.
#1+lnx+lny=(3^x)/(3^y)#

Differentiate left side

#1/x+1/y*dy/dx#

Differentiate right side using quotient rule

#(ln(3)*3^x*3^y-ln(3)*3^x*3^y*dy/dx)/((3^y)^2)#

Set equal again
#1/x+1/y*dy/dx = (ln(3)*3^x*3^y-ln(3)*3^x*3^y*dy/dx)/((3^y)^2)#

Solve for #dy/dx#

#dy/dx=((ln(3)*3^(x+y))/(3^(2y))-1/x)/(((ln(3)*3^(x+y))/3^(2y)+1/y)#