Question

What is the slope of the tangent line of `x^2-ye^(x-y)= C`, where C is an arbitrary constant, at `(0,0)`?

Answer 1

0

Explanation:

`x^2-ye^x/e^y=C` `0^2-0e^0=C` `C=0`

`e^y(x^2)=ye^x`

`color(blue)(d/dx)(e^y(x^2))=color(blue)(d/dx)(ye^x)`

`e^y(2x)+e^y(dy/dx)(x^2)=ye^x+(dy/dx)e^x`

`2xe^y-ye^x=(dy/dx)(e^x-e^y(x^2))`

`dy/dx=(2xe^y-ye^x)/(e^x-e^y(x^2)`

`dy/dx(0,0)=(0e^0-0e^0)/(e^0-0e^0=0/1=0`