`tan^-1(x^2 y) = x + xy^2` Find `(dy/dx)` by implicit differentiation.

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Chapter 3, 3.5 - Problem 17 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`tan^-1(x^2y)=x+xy^2`

taking derivative on both the sides

`d/(dx) tan^-1(x^2y)=d/(dx) (x+xy^2)`

`1/(1+(x^2y)^2) * d/(dx) (x^2y)=1+x(2y)dy/dx +y^2`

`1/(1+x^4y^2) *(x^2 dy/dx+y(2x)) =1+2xy dy/dx +y^2`

`(x^2/(1+x^4y^2)) dy/dx +(2xy)/(1+x^4y^2) = 1+ 2xy dy/dx + y^2`

`((x^2/(1+x^4y^2)) -2xy) dy/dx = 1+y^2 -(2xy)/(1+x^4y^2)`

`((x^2-2xy-2x^5y^3)/(1+x^4y^2)) dy/dx = (1+x^4y^2+y^2+x^4y^4-2xy)/(1+x^4y^2)`

`dy/dx = (1+y^2-2xy+x^4y^2+x^4y^4)/(x^2-2xy-2x^5y^3)`

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