Question

If e^(xy) = 2, then what is dy/dx at the point (1, ln2)

Answer 1

By implicit differentiation: 

(x(dy/dx) + y)e^(xy) = 0 

Note that when differentiating e^(xy), apply chain rule. When differentiating xy, use product rule. Also: When differentiating y w/respect to x, think of that as if you are differentiating f(x). 

Then, substitute (1,ln(2)) and solve for dy/dx. 

(1(dy/dx) + ln(2))e^(1ln(2)) = 0 
((dy/dx) + ln(2))e^(ln(2)) = 0 

Note that e^(ln(2)) = 2 since e and ln are inverse of each other. 

2((dy/dx) + ln(2)) = 0 
dy/dx + ln(2) = 0 . . . . You get this expression by dividing both sides by 2 
dy/dx = -ln(2) . . . . . . .Subtract both sides by ln(2) 

Therefore, dy/dx = -ln(2) 

I hope this helps!