Derivatives: prove there is no horizontal tangent line to graph


Differentiate implicitly:

$$2x-3y-3xy'+2yy'=0\implies(2y-3x)y'=3y-2x\implies y'=\frac{3y-2x}{2y-3x}$$

Thus, the only point where the derivative is defined and equals zero is when $\;3y-2x=0\;,\;\;(x,y)\in\text{ Dom}\,(f)\;$ , but then

$$y=\frac23x\implies f\left(x,\,\frac23x\right)=x^2-2x^2+\frac49x^2=1\iff-\frac59x^2=1$$

and this last equality is impossible (in the real number, of course), thus $\;y'\neq0\;$ for all points where it is defined.


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Bob Umadbro
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Bob Umadbro

Updated on August 16, 2022


  • Bob Umadbro
    Bob Umadbro about 1 year

    Equation: $-3xy + x^2 + y^2 = 1$

    Demonstrate that there is no horizontal tangent using inverse variation.

  • user247327
    user247327 about 7 years
    You mean, of course, that "$y'\ne 0$ for all points where it is defined".
  • DonAntonio
    DonAntonio about 7 years
    @user247327 Yes, of course. Thanks.