# Use Implicit differentiation to find the equation of the tangent line to the curve at the given point. `y^2-2x^3=-x^4 , (1,1)`

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We want to find tangent at point (1,1) on the curve

`y^2-2x^3=-x^4` (i)

Differentiate (i) with respect to x ( infac implicit differentiation)

`2y(dy)/(dx)-6x^2=-4x^3`

`(dy)/(dx)=(6x^2-4x^3)/(2y)`

`(dy)/(dx)=(3x^2-2x^2)/y`

Since derivative represent slope ,so we can get slope of the tangent at (1,1)

`((dy)/(dx))_{(1,1)}=(3xx 1^2-2 xx1^2)/1`

`` `=1`

Apply point slope equation to find an equation of tangent

`y-1=1(x-1)`

`y=x`

Thus required equation of the tangent at point (1,1) on the given curve is

`y=x`

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