Question

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

Answer 1

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}`

Answer 2

`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)`