Skip to main content ?

Section 3.10 : Implicit Differentiation

10. Find the equation of the tangent line to \({x^4} + {y^2} = 3\) at \(\left( {1, - \sqrt 2 } \right)\).

Show All Steps Hide All Steps

Start Solution

The first thing to do is use implicit differentiation to find \(y'\) for this function.

\[4{x^3} + 2y\,y' = 0\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\,\underline {y' = - \frac{{2{x^3}}}{y}} \]
Show Step 2

Evaluating the derivative at the point in question to get the slope of the tangent line gives,

\[m = {\left. {y'} \right|_{x = 1,\,\,y = - \sqrt 2 }} = - \frac{2}{{ - \sqrt 2 }} = \sqrt 2 \]
Show Step 3

Now, we just need to write down the equation of the tangent line.

\[y - \left( { - \sqrt 2 } \right) = \sqrt 2 \left( {x - 1} \right)\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\,\,y = \sqrt 2 \left( {x - 1} \right) - \sqrt 2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt 2 \left( {x - 2} \right)}}\]