지수함수와 로그함수의 미분

When we say a function is differentiable without specifying a range, we mean that it is differentiable over its domain. The function $f(x)$ is differentiable when $x \neq 1$ for any values of $a$ and $b$; it is up to us to figure out which constants make it differentiable when $x=1$.

In order to be differentiable, a function must be continuous. The definition of continuity tells us that, for $f$ to be continuous at $x=1$, we need $\displaystyle \lim_{x \to 1}f(x)=f(1)$. From the definition of $f$, we see $f(1)=a+b=\displaystyle \lim_{x \to 1^-}f(x)$, so we need $\displaystyle \lim_{x\to 1^+}f(x)=a+b$. Since $\displaystyle \lim_{x \to 1^+}f(x)=e^1=e$, we specifically need $$e=a+b.$$

Now, let's consider differentiability of $f$ at $x=1$. We need the following limit to exist:

$$\begin{aligned} \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}& \\ \text{In particular, we need the one-sided limits to exist and be equal:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h} \\ \text{If $h<0$, then $1+h<1$, so $f(1+h)=a(1+h)^2+b$. If $h>0$, then $1+h>1$, so $f(1+h)=e^{1+h}$. With this in mind, we begin to evaluate the one-sided limits:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^-}\frac{[a(1+h)^2+b]-[a+b]}{h}\\ &=\lim_{h \to 0^-}\frac{ah^2+2ah}{h}=2a\\ \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{e^{1+h}-(a+b)}{h} \\ \text{Since we take $a+b$ to be equal to $e$ (to ensure continuity):} \\ &= \lim_{h \to 0^+}\frac{e^{1+h}-e^1}{h}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}\right|_{x=1}=e^1=e \end{aligned}$$

So, we also need $$2a= e$$

Therefore, the values of $a$ and $b$ that make $f$ differentiable are $a=b=\dfrac{e}{2}$.