Showing that derivative of conjugate is conjugate of derivative, using chain rule

I'm trying to show that the derivative of the conjugate is the conjugate of the derivative, i.e. $$ \dfrac{\mathrm{d}[f(x)^*]}{\mathrm{d}x} = \biggl[\frac{\mathrm{d}f(x)}{\mathrm{d}x}\biggr]^*, $$ using the chain rule.

Calling the conjugate * function '\operatorname{conj}', we have by chain rule $$ \dfrac{\mathrm{d}\operatorname{conj}(f(x))}{\mathrm{d}x} = \dfrac{\mathrm{d}\operatorname{conj}(f(x))}{\mathrm{d}[f(x)]} \cdot \dfrac{\mathrm{d}f(x)}{\mathrm{d}x} $$

Now, $$ \dfrac{\mathrm{d}\operatorname{conj}(f(x))}{\mathrm{d}[f(x)]} \equiv \lim \limits_{h \to 0}{\frac{\operatorname{conj}(f(x)+h)-\operatorname{conj}(f(x))}{h}} = \lim \limits_{h \to 0}{\frac{h^*}{h}} $$

where $h \in C$.

I wasn't sure how to evaluate that limit, but according to Wolfram Alpha, it is = 1.

But that doesn't make sense, since then $$ \dfrac{\mathrm{d}\operatorname{conj}(f(x))}{\mathrm{d}x} = 1 \cdot \dfrac{\mathrm{d}f(x)}{\mathrm{d}x} = \dfrac{\mathrm{d}f(x)}{\mathrm{d}x}, $$ when it should be $\biggl[\dfrac{\mathrm{d}f(x)}{\mathrm{d}x}\biggr]^*$.

What am I doing wrong?

Solution 1:

So, $f$ is a map from $\mathbb R$ to $\mathbb C$ (which should have been made clear from the beginning). Suppose $f$ is differentiable at a point $x$, and $f'(x)=a$. This means exactly that $$\lim_{h\to 0}\frac{|f(x+h)-f(x)-ah|}{h} = 0 \tag1$$ Recalling that $|z|=|\bar z|$ for every complex number, we conclude from (1) that $$\lim_{h\to 0}\frac{\left|\overline{f(x+h)}-\overline{f(x)}-\bar a h\right|}{h} = 0 \tag2$$ (No need for bar over $h$, since $h$ is real.) Equation (2) says precisely that $\bar f$ has derivative $\bar a$ at $x$.

Generally, approaching derivatives via (1) rather than $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=a$$ is a good practice that will pay off in multivariable calculus.