Determine the equation(s) of tangent(s) line to the curve `y=4x^3-3x+5` which are perpendicular to the line `9y+x+3=0`

Explanation

Let `(x_{1}, y_{1})` be a point on the curve where we need to find the tangent(s).

Slope of the given line `=frac{-1}{9}`

Since, tangent is perpendicular to the given line,

Slope of the tangent `=frac{-1}{(frac{-1}{9})}=9`

Let `(x_{1}, y_{1})` be the point where the tangent is drawn to this curve.

Since, the point lies on the curve.

Hence,`y_{1}=4 x_{1}^{3}-3 x_{1}+5`

Now,`y=4 x^{3}-3 x+5`

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