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)