I am having trouble understanding the definition Hermitian and Hermitian conjugate.
An operator is Hermitian provided that: $\hat{O}^\dagger=\hat{O}$
The Hermitian conjugate of the differentiation operator: $\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^\dagger=-\frac{\mathrm{d}}{\mathrm{d}x}$
We know that the differentiation operator is not a Hermitian then why is it called a Hermitian conjugate and not just a conjugate
Hermitian is an adjective used to describe an operator which is equal to its Hermitian conjugate.
Hermitian conjugate (sometimes also called Hermitian adjoint) is a noun referring to the generalisation of the conjugate transpose of a matrix.
It doesn't really make sense to say that a particular operator is a Hermitian conjugate without any context. In your example, we would say that $-\frac{\mathrm{d}}{\mathrm{d}x}$ is the Hermitian conjugate of $\frac{\mathrm{d}}{\mathrm{d}x}$.
The Hermitian Conjugate or Hermitian Transpose of an operator $\hat{O}$ is defined as $\hat{O}^\dagger$.
As you stated in your question an operator $\hat{Q}$ is Hermitian iff $\hat{Q}=\hat{Q}^\dagger$, I know the terminology can be confusing. An operator is Hermitian if it is equal to its Hermitian Conjugate.
Now to the differentiation operator: I assume you know why $\frac{d}{dx}^\dagger=-\frac{d}{dx}$ since its in your question but here is a link that describes it if you are unsure: Explaining why $\mathrm{ d/d}x$ is not Hermitian, but $\mathrm{i~ d/d}x$ is Hermitian
Clearly $\frac{d}{dx}^\dagger \neq \frac{d}{dx}$, so the operator is $\textbf{not}$ Hermitian, but we can still just find its Hermitian Conjugate same as any other operator.
