Let $\mathcal{R}$ be an expansion of $(\mathbb{R},+,\cdot,-,<,0,1)$, and suppose that the exponential map is definable. I am asked to show that it is definable without parameters using the fact that the exponential is the unique differentiable function that verifies: $f'=f$ and $f(0)=1$.
I can express the fact that for all $x\in\mathbb{R}$ $f'(x)=f(x)$ and $f(0)=1$ by a formula involving the symbol $f$. But I dont know how to use this to show the claim.
Edit(after @Noah Schweber response)
First note that the absolute value is definable by the formula $\varphi_{|\cdot|}(h,h')$ : $(0<h\land h'=h)\lor(\neg(0<h)\land h'=-h)$
next we have \begin{align*} f(0)=1\wedge f'=f &\Leftrightarrow f(0)=1 \wedge \forall a \ f'(a)=f(a)\\ &\Leftrightarrow f(0)=1 \wedge \ \forall a \ \forall\epsilon>0 \ \exists\eta>0 \ \forall h(0<|h|<\eta\rightarrow \\ &|\frac{f(a+h)-f(a)}{h}-f(a)|<\epsilon)\\ &\Leftrightarrow f(0)=1 \wedge \ \forall a \ \forall\epsilon>0 \ \exists\eta>0 \ \forall h(0<|h|<\eta\rightarrow \\ &|f(a+h)-f(a)-hf(a)|<|h|\epsilon)\\ &\Leftrightarrow f(0)=1 \wedge \ \forall a \ \forall\epsilon>0 \ \exists\eta>0 \ \forall h(0<|h|<\eta\rightarrow \\ &|f(a+h)-(1+h)f(a)|<|h|\epsilon)\\ &\Leftrightarrow f(0)=1 \wedge \ \forall a \ \forall\epsilon>0 \ \exists\eta>0 \ \forall h(\exists h' \ \varphi_{|\cdot|}(h,h') \ (0<h'<\eta\rightarrow \\ &-h'\epsilon<f(a+h)-(1+h)f(a)<h'\epsilon)) \end{align*} In the last sentence we need to replace for example $\forall\epsilon>0\ldots$ by $\forall\epsilon[\epsilon>0\rightarrow\cdots]$ and so on.
Now if the exponential is definable in $\mathcal{R}$ by a formula $\psi(x,y,\bar{a})$ then it is parameter-free definable by $\xi(x,y)$ : $$\exists \bar{z}(\psi(0,1,\bar{z}) \wedge \ \forall a \ \forall\epsilon>0 \ \exists\eta>0 \ \forall h(\exists h'\exists s,t \ \varphi_{|\cdot|}(h,h') \land \psi(a,s,\bar{z})\land \psi(a+h,t,\bar{z}) \ (0<h'<\eta\rightarrow -h'\epsilon<t-(1+h)s<h'\epsilon))\land \psi(x,y,\bar{z}))$$ Am I right?
The point is that the exponential function is "recognizable" in $\mathbb{R}$ already: via the differential equation, we can write a formula $\varphi$ in the language of $\mathbb{R}$ plus a new function symbol $\dot{f}$ such that $(\mathbb{R},f)\models\varphi$ iff $f$ is the exponential function.
Now suppose the exponential function is definable-with-parameters in $\mathcal{R}$ as $\psi(x,y,\overline{a})$. Then the set of tuples $\overline{b}$ such that the function defined by $\psi(x,y,\overline{b})$ satisfies $\varphi$ (= is the exponential function) is definable without parameters via some formula $\theta(\overline{z})$. From this, the exponential function itself is definable without parameters: $e^x=y$ iff $\exists\overline{z}(\theta(\overline{z})\wedge\psi(x,y,\overline{z}))$.
More generally, whenever a function is "implicitly definable" over a structure $\mathcal{M}$, it is parameter-freely definable in any expansion of $\mathcal{M}$ in which it is definable with parameters. (Note that there is some ambiguity here around the phrase "implicitly definable," namely whether we look at a theory or a specific structure; see also here.)