determine the equation of the tangent/y-e^(xy)=0 at A(0,1)

G

Guest

Guest
For each of the following, determine the equation of the tangent at the given point.

For the curve defined by y-e^(xy)=0 at A(0,1)

Im having trouble doing this since you have to se implicit differentiation right?
also can you bring the 1 to one side like

y'= dy/dx-e^xy(1y+x(dy/dx)
dy/dx-e^(xy)y- e^(xy) x(dy/dx)
e^(xy) Y=dy/dx (1-e^xy) X
dy/dx= e^(xy) Y/[1-e^xy)X

Im not sure if I did this right..and I have no idea how to carry on,heeeeeeeeelpp please..thanks so much!!
 
\(\displaystyle \L
y - e^{xy} = 0\quad \Rightarrow \quad \frac{{dy}}{{dx}} - \left( {y + x\frac{{dy}}{{dx}}} \right)e^{xy} = 0\)
 
bittersweet said:
but how do you find the equation??
Click to expand...
Gee, you have got to do somethings for yourself!
Let x=0 & y=1.
Solve for \(\displaystyle \frac{{dy}}{{dx}}.\)
 
You must log in or register to reply here.
Top