Tan^-1 (x^2y)=x+xy^2

Question

Need to use implicit differentiation.

Answer 1

When we use implicit differentiation, we derive both sides of the equation using the chain rule.  I will use the notation y' to indicate derivative.

 

Recall the derivative of tan^-1(x) = 1 / (1 + x²)

 

 

d/dx[tan^-1(x²y)] = d/dx[x + xy²]

 

(2xy + x²y') / (1 + x⁴y²) = 1 + y² + 2xyy'

 

 

Multiply both sides of the equation by the left side's denominator .

 

2xy + x²y' = (1 + x⁴y²)(1 + y² + 2xyy')

 

2xy + x²y' = 1(1 + y² + 2xyy') + x⁴y² (1 + y² + 2xyy')

 

2xy + x²y' = 1 + y² + 2xyy' + x⁴y² + x⁴y⁴ + 2x⁵y³y'

 

 

Now that the hard part is done, we move all the y' terms to the left side of equation and all the non-y' terms to the right side of equation.

 

x²y' - 2xyy' - 2x⁵y³y' = 1 - 2xy + y² + x⁴y² + x⁴y⁴ 

 

 

Factor out the y' on the left side of equation.

 

y' (x² - 2xy - 2x⁵y³) = 1 - 2xy + y² + x⁴y² + x⁴y⁴ 

 

 

Finally, divide both sides of the equation by the coefficient of y'.