tan^3 (xy^2 + y)=x

Question

 find dy/dx

Answer 1

We must use our chain rule on the left side of the equation:

3*tan(xy²+y))² * (x*2y(dy/dx)+y² + (dy/dx)) = 1

 

Note: (2xy(dy/dx)+y² + (dy/dx)) is the derivative of the inside and (2xy(dy/dx)+y²) is our product rule!

 

Now we want to get dy/dx by itself, so we can first divide by 3tan(xy²+y)²

(2xy(dy/dx)+y² + (dy/dx)) = 1/(3tan(xy²+y)²)

 

Now subtract y² from both sides:

2xy(dy/dx) + (dy/dx) = 1/(3tan(xy²+y)²) - y²

 

Now factor out the dy/dx from the left side:

(dy/dx)(2xy+1) = 1/(3tan(xy²+y)²) - y²

 

Divide both sides by 2xy+1:

(dy/dx) =  1/(3(2xy+1)*tan(xy²+y)²) - y²/(2xy+1)