I want to solve the following exercise:
Let $\mathcal{C}\subseteq P^{2}(\mathbb{C})$ be an irreducible curve of degree $n\geq1$. Let $m_P (\mathcal C)$ be the multiplicity of $P$, prove that $$\sum_{P}m_{P}(\mathcal{C})(m_{P}(\mathcal{C})-1)\leq n(n-1)$$and prove that $\mathcal{C}$ has at most $\frac{n(n-1)}{2}$ singular points.
I think I can use this somehow, but maybe I am wrong:
Let $\mathcal C$ and $\mathcal C'$ be two curves, then: $$\sum_{P}m_{P}(\mathcal{C})m_{P}(\mathcal{C}')\leq \mathrm{degree}(\mathcal C)\cdot \mathrm{degree}(\mathcal C ')$$
My intuition tells me that if $\mathcal C =\mathrm V (F)$, then maybe I can take $\mathcal C'=\mathrm V (F')$, where $F'$ is some derivative of $F$, so the multiplicity of P in the new curve will be the multiplicity in $\mathcal C$ less one, and the degree of the new curve will be $n-1$.
How can I take that derivative? Would any partial derivative work here? Is it true that, for example $\mathcal C'=\mathrm V \left( \frac{\partial F}{\partial X}\right)$ verifies this?
Also, I don't know how to answer the second part.
As you suspected, you should pick $\mathcal C'$ to be a polar curve of $\mathcal C$. You cannot always choose $F_X$ as the defining equation because it might be zero, but some linear combination of the partial derivatives will work.
For the second part, note that $P$ is a singular point of $\mathcal C$ if and only if $m_P(\mathcal C)\ge 2$ and the number $k$ of singular points of $\mathcal C$ can therefore be estimated as $$ k = \#\{P\mid m_P(\mathcal C)>1\} \le \tfrac12 \sum_P 2\cdot (m_P(\mathcal C)-1)\le \tfrac12 \sum_{P} m_P(\mathcal C)\cdot(m_P(\mathcal C)-1) \le \tfrac{n(n-1)}2. $$