# Why is derivative of $x$ with respect to $x$ equal to $1$?

I just started learning the derivatives of inverse function. The first example is based on the fact that $\frac{\mathsf{d}x}{\mathsf{d}x} = 1$ and it is stated that I should know this already. However I don't recall this fact, moreover I don't know how you can solve $\frac{\mathsf{d}x}{\mathsf{d}x}$.

Could somebody explain me how do we solve such question and is this rule apply to every function?

• $\mathrm d x$ is a constant quantity, at least from the point of view of non-standard analysis. But $\mathrm d$ alone is an operator. The notation is slightly confusing. – Masacroso Feb 19 '16 at 18:09
• @Masacroso It doesn't seem like a good idea to start learning about derivatives with nonstandard analysis if standard analysis was not even learned ;-) – AlexR Feb 19 '16 at 18:13
• Maybe @AlexR, Im not sure. Many people says that non-standard analysis, not in the complex set theory behind, is easier to approach that the standard one. And let me point too that the notation $\mathrm d x$ is Leibniz notation, and non-standard analysis is the continuation of Leibniz "materialistic" math. This is one of the reasons why the notation is not completely consistent with the standard analysis. Behind a symbolism is a mind, in this case Leibniz mind. – Masacroso Feb 19 '16 at 20:54

Maybe it becomes clear if you write it as $$\frac{\mathrm d}{\mathrm dx} x = (x)' = 1$$ This is basically the first derivative you learn of.
• @KimPeek The first class of derivatives I learned about was in fact that of (affine) linear functions of the form $f(x) = cx+b$. Indeed more people in my class had trouble seeing that the derivative of a constant was $0$ than seeing that the derivative of $cx$ was $c$. – AlexR Feb 19 '16 at 18:12
$$\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ which in this case ($f(x)=x$) reduces to $$\lim_{h \rightarrow 0}\frac{x+h-x}{h}=\lim_{h \rightarrow 0}\frac{h}{h}=1$$