My course book on differential calculus says, "... the equation of the tangent at the origin can be written down by equating to zero the lowest degree terms in x and y." I don't see how this is true.

Why is this true ?

  • 2
    $\begingroup$ '' the lowest degree terms'' of what? $\endgroup$ Jul 12, 2017 at 15:11
  • $\begingroup$ Sorry. I've edited the question now. $\endgroup$
    – Bibek_G
    Jul 12, 2017 at 15:34
  • $\begingroup$ A term to search for is "tangent cone". $\endgroup$ Jul 12, 2017 at 16:10

1 Answer 1


Let the curve be $$f(x)=g(y)$$

Assuming that $f(\cdot), g(\cdot)$ are polynomials, the respective constant terms must be zero such that the curve passes through the origin.

Differentiating w.r.t $x$ gives $$f'(x)+g'(y)\frac {dy}{dx}=0\\\frac {dy}{dx}=-\frac {f'(x)}{g'(y)}$$ Hence slope at origin is $-\dfrac{f'(0)}{g'(0)}$ and tangent at origin is $$y=-\frac {f'(0)}{g'(0)}x\\ \color{red}{f'(0)\ x+g'(0)\ y=0}$$


If $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0$, then $f'(0)\ x = a_1 x$ which is the lowest term in $x$ in $f(x)$. Similarly for $g(y)$.

Instruction should read
"...by equating to zero the sum of the lowest degree terms in $x$ and $y$."

Note that, assuming $g'(0)\neq 0$, if $f(x)$ does not have an $x$ term then the tangent at the origin must have a slope of zero i.e. the $x$-axis is the tangent at the origin.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .