Question

How do you implicitly differentiate `ysinx^2 = xsin(e^y)`?

Answer 1

`dy/dx`= `(sin(e^y)-2xycos(x^2))/(sin(x^2)-e^yxcos(e^y))`

Explanation:

`y times 2xcos(x^2) + sin(x^2) times dy/dx` = `x times e^ycos(e^y) times dy/dx + sin(e^y)`

`2xycos(x^2) + dy/dx times sin(x^2)` = `dy/dx times e^yxcos(e^y) + sin (e^y)`

`dy/dx sin (x^2) - e^ycos(e^y)` = `sin (e^y) - 2xycos(x^2)`

`dy/dx` = `(sin(e^y)-2xycos(x^2))/(sin(x^2)-e^yxcos(e^y))`